Have you ever thought about how we calculate the distance travelled by a vehicle with changing speed, or find the exact area of an irregular shape like a curved field? Application of Integrals provides the answer to such real-life problems. In NCERT Exemplar Class 12 Maths Chapter 8 Application of Integrals, students learn how integrals are used to find the area under a curve and between curves, which helps in calculating quantities related to motion, work, and space.
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NCERT Exemplar Class 12 Maths Solutions Chapter 8 Application Of Integrals
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In simple terms, the Application of Integrals means using integration to solve practical problems where values change continuously. This chapter plays an important role in understanding real-world applications in physics, economics, and engineering. Students are advised to practise NCERT Exemplar Class 12 Maths Solutions regularly, as it strengthens conceptual understanding and improves problem-solving skills. Check this NCERT article for complete syllabus coverage along with NCERT Books, Solutions, Syllabus, and Exemplar Problems with Solutions.
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NCERT Exemplar Class 12 Maths Solutions Chapter 8 Application Of Integrals
| Class 12 Maths Chapter 8 Exemplar Solutions Exercise: 8.3 Page number: 176-178 Total Questions: 34 |
Question 1
Find the area of the region bounded by the curves $y^2 = 9x, y = 3x.$
Answer:
It is mentioned in the question that,
$y^{2} = 9x$ belongs to a parabola and $y = 3x$ belong to a straight line which passes through origin.
Starting with a rough figure showing those equations below,
From the parabola equations, it is seen that x cannot be negative, so the graph would be on the right of the X-axis. The parabola would thus be opening to the right.
To find the points of the two equations mentioned, you have to solve the two equations simultaneously.
$\begin{array}{l} \text { Put } y=3 x \text { in } y^{2}=9 x \\ \Rightarrow(3 x)^{2}=9 x \\ \Rightarrow 9 x^{2}=9 x \\ \Rightarrow x^{2}-x=0 \\ \Rightarrow x(x-1)=0 \\ \Rightarrow x=0 \text { and } x=1 \end{array}$
We got the coordinates of x, to find the coordinates of y.
Put x = 1 and x = 0 in the equation of y = 3x.
It is found that the y coordinates are y = 3 and y = 0, respectively.
Therefore, the point of intersection of the parabola and the straight line is (1, 3) and (0, 0)
Now, calculate the area enclosed between the parabola and the straight line.
For calculating the area, we have to subtract the area under the straight line, which extends from x = 0 to x = 1, from the area under the parabola.
It can be written as,
Area between parabola and straight line = area under parabola – area under straight line …. (1)
On calculating the area under the parabola,
$y^{2 = }9x \\$
$\int y = 3 \sqrt x\\$
On integrating the above equation from 0 to 1
$\\ \Rightarrow \int_{0}^{1} y d x=3 \int_{0}^{1} \sqrt{x} d x \\ \Rightarrow \int_{0}^{1} y d x=3 \int_{0}^{1} x^{\frac{1}{2}} d x \\ \Rightarrow \int_{0}^{1} y d x=3\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{1} \\ \Rightarrow \int_{0}^{1} y d x=3\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1} \\$
$\\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=3 \frac{2}{3}\left[\mathrm{x}^{\frac{3}{2}}\right]_{0}^{1} \\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=2\left[1^{\frac{3}{2}}-0\right] \\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=2$
Now, for calculating the area under line y = 3x i.e. the area of triangle OAB
$\int y = 3x \\$
On integrating the above equation from 0 to 1
$\\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=\int_{0}^{1} 3 \mathrm{xdx} \\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=3\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{1} \\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=3\left(\frac{1^{2}}{2}-0\right) \\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=\frac{3}{2}$
Using the equation mentioned above (1)
area between parabola and straight line $= 2 - \frac{3}{2} = 1/2\ unit^{2 }\\$
Therefore,
It is found that the area was 1/2 unit2 between the curves $y^2 = 9x\: \: \: and\: \: \: y = 3x.$
Question 2
Find the area of the region bounded by the parabola $y^2 = 2px, x^2 = 2py$
Answer:
The equation of the parabola is $y^{2 }= 2px$, which shows no negative values for x. Therefore, the graph will be on the right of the Y-axis and will pass through (0, 0)
Similarly, for equations $x^{2} = 2py$, there are no negative values for y, therefore, the graph will be above the X axis passing through (0, 0).
To find a point of interaction, let us solve the equations simultaneously.
$\\ \text { Put } y=\frac{x^{2}}{2 p} \text { in } y^{2}=2 p x \\ \Rightarrow\left(\frac{x^{2}}{2 p}\right)^{2}=2 p x \\ \Rightarrow \frac{x^{4}}{4 p^{2}}=2 p x \\ \Rightarrow x^{4}=8 p^{3} x \\ \Rightarrow x^{3}=8 p^{3} \\$
$\\ \Rightarrow x=2 p \\ \text { Put } x=2 p \text { in } y^{2}=2 p x \\ \Rightarrow y^{2}=2 p(2 p) \\ \Rightarrow y=2 p$
The interaction point was found to be (2p, 2p)
Now, we have to find out the areas between the two parabolas.
The equation can be written as
The area between the two parabolas = area under the parabola
$y^{2 }= 2px - \text{area under the parabola } x^{2 }= 2py \dots (1) \\$
For finding the area under$y^{2 }= 2px\: \: parabola \\$
$\int y = \sqrt 2p \sqrt x \\$
Integrating the equation from 0 to 2p
$\\ \Rightarrow \int_{0}^{2 p} y d x=\sqrt{2 p} \int_{0}^{2 p} \sqrt{x} d x \\ \Rightarrow \int_{0}^{2 p} y d x=\sqrt{2 p} \int_{0}^{2 p} x^{\frac{1}{2}} d x \\ \Rightarrow \int_{0}^{2 p} y d x=\sqrt{2 p}\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{2 p} \\ \Rightarrow \int_{0}^{2 p} y d x=\sqrt{2 p}\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{2 p} \\ \Rightarrow \int_{0}^{2 p} y d x=\sqrt{2 p} \frac{2}{3}\left[(2 p)^{\frac{3}{2}}-0\right]$
$\\\Rightarrow \int_{0}^{2 p} y d x=\frac{2}{3}(2 p)^{\frac{1}{2}}(2 p)^{\frac{3}{2}} \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{2}{3}(2 p)^{\frac{1}{2}+\frac{3}{2}} \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{2}{3}(2 p)^{2} \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{8 p^{2}}{3}$
For finding the area under $x^{2 }= 2py \text{ parabola} \\$
$\int y = x^{2}/ 2p \\$
Integrating from 0 to 2p
$\\\Rightarrow \int_{0}^{2 p} y d x=\int_{0}^{2 p} \frac{x^{2}}{2 p} d x \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{1}{2 p}\left[\frac{x^{3}}{3}\right]_{0}^{2 p} \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{1}{2 p}\left[\frac{(2 p)^{3}}{3}\right] \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{(2 p)^{2}}{3} \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{4 p^{2}}{3}$
Using
(i)
$\\\Rightarrow$ area bounded by two parabolas given $=\frac{8 \mathrm{p}^{2}}{3}-\frac{4 \mathrm{p}^{2}}{3}$ $\\\Rightarrow$ area bounded by two parabolas given $=\frac{4 \mathrm{p}^{2}}{3}$
Hence area is $\frac{4 \mathrm{p}^{2}}{3}$ unit ${ }^{2}$
Question 3
Find the area of the region bounded by the curve $y = x^3$ and y = x + 6 and x = 0.
Answer:
Let's start with a rough plot of the curve $y=x^3$ along with the lines $\mathrm{y}=\mathrm{x}+6$ and $\mathrm{x}=0$ When $\mathrm{X}=0$, it means Y axis
The questions say to find the area between the curve and the line and Y axis
First solve the $y=x+6$ and $y=x^3$, in order to find the interaction point
$
\begin{aligned}
& \text { Put } y=x^3 \text { in } y=x+6 \\
& \Rightarrow x^3=x+6 \\
& \Rightarrow x^3-x-6=0
\end{aligned}
$
For checking $0,1,2$ satisfies this cubic, showing 2 is one factor, therefore $x-2$ is a factor.
Solving the equation,
$
\Rightarrow(x-2)\left(x^2+2 x+3\right)=0
$
Observe that $x^2+2 x+3$ doesn't have real roots
Therefore, $\mathrm{x}=2$
Substituting this $x=2$ in $y=x+6, y=8$
Therefore, curves intersect at $(2,8)$
The area bounded will be
Area bounded $=$ area by $y=x^3$ on Y axis - area by $\mathrm{y}=\mathrm{x}+6$ on Y axis $\ldots$
For finding the area under $y=x^3$
$
\Rightarrow x=\sqrt[3]{y}
$
Integrate the equation from 0 to 8
$
\begin{aligned}
& \Rightarrow \int_0^8 \mathrm{xdy}=\int_0^8 \mathrm{y} \frac{1}{3} \mathrm{dy} \\
& \Rightarrow \int_0^8 \mathrm{xdy}=\left[\frac{\mathrm{y}^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right]_0^8 \Rightarrow \int_0^8 x d y=\frac{3}{4}\left(2^3\right)^{\frac{4}{3}} \\
& \Rightarrow \int_0^8 \mathrm{xdy}=\left[\frac{\mathrm{y}^{\frac{4}{3}}}{\frac{4}{3}}\right]_0^8 \Rightarrow \int_0^8 x d y=\frac{3}{4} 2^4 \\
& \Rightarrow \int_0^8 \mathrm{xdy}=\frac{3}{4}\left[\mathrm{y}^{\frac{4}{3}}\right]_0^8 \Rightarrow \int_0^8 x d y=\frac{3}{4} 16 \\
& \Rightarrow \int_0^8 \mathrm{xdy}=\frac{3}{4}[8 \overline{4 / 3}-0]
\end{aligned}
$
Now, finding the area under $y=x+6$
For finding the area from 6 to 8, because the line passes through the $Y$ axis at 6 and extends up to 8, the point where the curve and line intersect
$
\text { item } X=y-6
$
Integrating from 6 to 8
$
\Rightarrow \int_6^8 x d y=\int_6^8(y-6) d y
$
$
\begin{aligned}
& \Rightarrow \int_6^8 x d y=\left[\left(\frac{8^2}{2}-6(8)\right)-\left(\frac{6^2}{2}-6(6)\right)\right] \\
& \Rightarrow \int_6^8 x d y=[32-48-18+36] \\
& \Rightarrow \int_6^8 x d y=2
\end{aligned}
$
Using equation (1)
Area bound was found to be 12-2 = $10 \mathrm{unit}^2$
Therefore, the area was found to be 10 unit $^2$
Question 4
Find the area of the region bounded by the curve $y^2 = 4x, x^2 = 4y.$
Answer:
The equation $y^{2 }= 4x$ is a parabola, and no negative values of x are seen; therefore, this parabola lies to the right of the Y axis passing through (0, 0)
Similarly, for $x^{2} = 4y$, which is a parabola, undefined negative values of y lie above the X-axis and pass through (0, 0)
To find a point of interaction, solve simultaneously.
$\begin{aligned} &\text { Put } y=\frac{x^{2}}{4} \text { in } y^{2}=4 x\\ &\Rightarrow\left(\frac{x^{2}}{4}\right)^{2}=4 x\\ &\Rightarrow \frac{x^{4}}{16}=4 x\\ &\Rightarrow x^{4}=64 x\\ &\Rightarrow x^{3}=64\\ &\Rightarrow x=4\\ &\text { Put } x=4 \text { in } y^{2}=4 x\\ &\Rightarrow y^{2}=4(4)\\ &\Rightarrow y=4 \end{aligned} \\$
The point of interaction is, therefore (4, 4)
For finding the area between two parabolas
Area between two parabolas = area under parabola $y^{2 }= 4x -\text{area under parabola } x^{2} = 4y \ldots . (1) \\$
Let us find area under parabola $y^{2}= 4x\\$
$= y = 2 \sqrt{x}$
Integrate from 0 to 4
$\Rightarrow \int_{0}^{4} \mathrm{ydx}=2 \int_{0}^{4} \sqrt{\mathrm{x}} \mathrm{dx} \\ \begin{aligned} &\Rightarrow \int_{0}^{4} \mathrm{ydx}=2 \int_{0}^{4} \mathrm{x} \frac{1}{2} \mathrm{dx}\\ &\Rightarrow \int_{0}^{4} \mathrm{ydx}=2\left[\frac{\mathrm{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{4}\\ &\Rightarrow \int_{0}^{4} \mathrm{ydx}=2\left[\frac{\mathrm{x} \frac{3}{2}}{\frac{3}{2}}\right]_{0}^{4}\\ &\Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{4}{3}\left[\mathrm{x}^{\frac{3}{2}}\right]_{0}^{4}\\ &\Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{4}{3}\left[4^{\frac{3}{2}}-0\right]\\ &\Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{4}{3}\left(2^{2}\right)^{\frac{3}{2}} \end{aligned} \\$
$\\ \Rightarrow \int_{0}^{4} y d x=\frac{4}{3}(2)^{3} \\ \Rightarrow \int_{0}^{4} y d x=\frac{32}{3} \\ \\$
To find area under parabola $x^{2}= 4y\\$
$= x^{2}= 4 y\\$
$\\\Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{4}{3}(2)^{3}\\$
$\Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{32}{3}\\ $
$\text{Now let us find the area under the parabola}$
$x^{2}=4 y\\ \Rightarrow x^{2}=4 y\\ \Rightarrow \mathrm{y}=\frac{\mathrm{x}^{2}}{4} \\$ $\text{Integrate from 0 to 4}\\$
$\Rightarrow \int_{0}^{4} \mathrm{ydx}=\int_{0}^{4} \frac{\mathrm{x}^{2}}{4} \mathrm{~d} \mathrm{x} \\$
$ \Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{1}{4}\left[\frac{\mathrm{x}^{3}}{3}\right]_{0}^{4} \\ $
$\Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{1}{4}\left[\frac{4^{3}}{3}\right] \\ \Rightarrow \int_{0}^{4} y d x=\frac{4^{2}}{3} \\ \Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{16}{3} \\ \text{Using (i)} \\$
$ \Rightarrow \text{area bounded by two parabolas given} =\frac{32}{3}-\frac{16}{3} \\$ $\Rightarrow \text{area bounded by two parabolas given} =\frac{16}{3} \\$
Therefore, the area is found to be $\frac{16}{3}$ unit $^2$
Question 5
Find the area of the region included between $y^2 = 9x$ and y = x
Answer:
The equation$y^{2} = 9x$ has no negative values of x.
Therefore, it lies on the right of the Y axis passing through (0, 0).
And y = x depicts a straight line through the origin.
To find the area, see the picture shown below.
To find the point of interaction, solve the two equations simultaneously.
$\\Put \: \: y=x \text{ in } y^{2}=9 x \Rightarrow x^{2}=9 x \Rightarrow x=9\\Put\: \: x=9 in y=x \text{ we get } y=9 \\$
Therefore, the point of interaction is (9, 9)
The area between the parabola and line = area under parabola – area under line $\ldots (1) \\$
Let us find the area under the parabola
$\\\Rightarrow y^{2}=9 x \\ \Rightarrow y=3 \sqrt{x} \\ \text{Integrate from 0 to 9}\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=\int_{0}^{9} 3 \mathrm{x}^{\frac{1}{2}} \mathrm{dx}\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=3\left[\frac{\mathrm{x} \frac{1}{2}+1}{\frac{1}{2}+1}\right]_{0}^{9}\\ \Rightarrow \int_{0}^{9} y \mathrm{dx}=3\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{9}\\ \\$
$\\\Rightarrow \int_{0}^{9} \mathrm{ydx}=3 \frac{2}{3}\left[9 \frac{3}{2}-0\right]\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=2\left(3^{2}\right)^{\frac{3}{2}}\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=2\left(3^{3}\right)\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=54\\$
Now let us find area under straight line $\mathrm{y}=\mathrm{x}$
y=x
Integrate from 0 to 9
$\\\Rightarrow \int_{0}^{9} \mathrm{ydx}=\int_{0}^{9} \mathrm{xdx} \\\Rightarrow \int_{0}^{9} \mathrm{ydx}=\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{9} \\\Rightarrow \int_{0}^{9} \mathrm{ydx}=\left(\frac{9^{2}}{2}-0\right)\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=\frac{81}{2}\\ \Rightarrow \int_{0}^{9} y \mathrm{dx}=40.5$
Using (i)
$\Rightarrow$ area between parabola and line $=54-40.5=13.5 unit ^{2}$
Therefore, the area was found to be $13.5 unit^{2}. \\$
Question 6
Find the area of the region enclosed by the parabola $x^2 = y$ and the line y = x + 2
Answer:
The equation $x^{2} = y$ depicts a parabola and has no negative values for y, therefore, it lies above the X-axis and passes through the origin (0, 0).
And y = x + 2 is a straight line.
The figure below shows the area to be found,
Let’s find the point of intersection of the two equations.
$\\\text{Put } y=x+2 inx^{2}=y\\ \Rightarrow x^{2}=x+2\\$
$ \Rightarrow x^{2}-x-2=0\\$$ \Rightarrow x^{2}-2 x+x-2=0\\ \Rightarrow x(x-2)+1(x-2)=0\\$
$ \Rightarrow(x+1)(x-2)=0\\$
$ \Rightarrow x=-1 \text{ and } x=2\\$
$ \text{Put } x=-1 \text{ and } x=2 in x^{2}=y \text{ we get } y=1 \text{ and } y=4 respectively \\$
The point of interaction was found to be (-1, 1) and (2, 4)
The area between the line and parabola = area under line – area under a parabola
Let us find the area under line y=x+2
$\\ \Rightarrow y=x+2\\ \text{Integrate from -1 to 2} \Rightarrow \int_{-1}^{2} y d x=\int_{-1}^{2}(x+2) d x\\$
$ \Rightarrow \int_{-1}^{2} y \mathrm{~d} x=\left[\frac{x^{2}}{2}+2 x\right]_{-1}^{2}\\$
$ \Rightarrow \int_{-1}^{2} y d x=\left[\left(\frac{2^{2}}{2}+2(2)\right)-\left(\frac{(-1)^{2}}{2}+2(-1)\right)\right]\\$
$ \Rightarrow \int_{-1}^{2} y d x=6-\left(\frac{1}{2}-2\right) \\$
$ \Rightarrow \int_{-1}^{2} \mathrm{ydx}=6-\left(\frac{-3}{2}\right)\\$
$ \Rightarrow \int_{-1}^{2} y \mathrm{dx}=6+\frac{3}{2}\\$$ \Rightarrow \int_{-1}^{2} \mathrm{y} \mathrm{dx}=\frac{15}{2}\\$
Now, let us find the area under the parabola
$\\ x^{2}=y \Rightarrow y=x^{2}\\ \text{Integrate from -1 to 2 }\\ \Rightarrow \int_{-1}^{2} y d x=\int_{-1}^{2} x^{2} d x \\$
$\\\begin{aligned} &\Rightarrow \int_{-1}^{2} \mathrm{ydx}=\left[\frac{\mathrm{x}^{3}}{3}\right]_{-1}^{2}\\ &\Rightarrow \int_{-1}^{2} \mathrm{ydx}=\left(\frac{2^{3}}{3}-\frac{(-1)^{3}}{3}\right)\\ &\Rightarrow \int_{-1}^{2} \mathrm{ydx}=\left(\frac{8}{3}+\frac{1}{3}\right)\\ &\Rightarrow \int_{-1}^{2} \mathrm{ydx}=3\\ &\text { Using (i) }\\ &\text { area enclosed by line and parabola }=\frac{15}{2}-3=\frac{9}{2} \text { unit }^{2} \end{aligned} \\$
Therefore, the area was found to be$\frac{9}{2}$ unit $^2$
Question 7
Find the area of region bounded by the line x = 2 and the parabola $y^2 = 8x$
Answer:
The equation $y^{2} = 8x$ is a parabola not defining negative values of x, therefore, it lies to the right of the Y axis passing through the origin.
And x = 2 is a straight line parallel to the Y-axis.
The figure below shows the area to be calculated.
We have to integrate $y^{2} = 8x \\$
$\\Y = 2 \sqrt 2 \sqrt x \text{ from 0 to 2} \\$
On integrating the above equation from 0 to 2, it would give the area enclosed under quadrant 1 only. Therefore, to find the area in quadrant 2, we have to multiply it by 2 since it is symmetrical.
Therefore the area ODBC = 2 x area OBC $\ldots .(1) \\$
Let us find the area under the parabola
$\\ \Rightarrow y^{2}=8 x\\ \Rightarrow y=2 \sqrt{2} \sqrt{x}\\ \text{Integrate from 0 to 2}\\ \Rightarrow \int_{0}^{2} y d x=\int_{0}^{2} 2 \sqrt{2} x^{\frac{1}{2}} d x\\ \Rightarrow \int_{0}^{2} \mathrm{ydx}=2 \sqrt{2}\left[\frac{\mathrm{x}^{ \frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{2}\\ \Rightarrow \int_{0}^{2} \mathrm{ydx}=2 \sqrt{2}\left[\frac{\mathrm{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{2} \\$
$\\\begin{aligned} &\Rightarrow \int_{0}^{2} \mathrm{ydx}=2 \sqrt{2} \frac{2}{3}\left[\mathrm{x}^{\frac{3}{2}}\right]_{0}^{2}\\ &\Rightarrow \int_{0}^{2} y d x=\frac{4}{3}\left(2^{\frac{1}{2}}\right)\left(2^{\frac{3}{2}}-0\right)\\ &\Rightarrow \int_{0}^{2} y d x=\frac{4}{3}\left(2^{\frac{1}{2}+\frac{3}{2}}\right)\\ &\Rightarrow \int_{0}^{2} y d x=\frac{4}{3}\left(2^{2}\right)\\ &\Rightarrow \int_{0}^{2} \mathrm{ydx}=\frac{16}{3}\\ &\Rightarrow \operatorname{areaOBC}=\frac{16}{3}\\ &\text { Using (i) } \end{aligned} \\$
The shaded areaOCBD $=2 \times \frac{16}{3}$
Hence area bounded $=\frac{32}{3}$ unit $^2$
Question 8
Answer:
$
y=\sqrt{4-x^2}
$
Square both sides
$
\begin{aligned}
& \Rightarrow y^2=4-x^2 \\
& \Rightarrow x^2+y^2=4 \\
& \Rightarrow x^2+y^2=2^2
\end{aligned}
$
The above equation depicts a circle with an origin and radius of 2
Equation of X axis is $\mathrm{y}=0$
Point of interaction are $(-2,0)$ and $(2,0)$
The figure below shows the area
Now, let us find the area
$y=\sqrt{4-x^{2}}$
Integrate from -2 to 2
$\Rightarrow \int_{-2}^{2} \mathrm{ydx}=\int_{-2}^{2} \sqrt{4-\mathrm{x}^{2}} \mathrm{~d} \mathrm{x} \\$
$\\\begin{aligned} &\text { Using uv rule of integration where u and } v \text { are functions of } x\\ &\int_{a}^{b} u v d x=\left[u \int \operatorname{vdx}\right]_{a}^{b}-\int_{a}^{b}\left(u^{\prime} \int v d x\right) d x\\ &\text { Here } u=\sqrt{4-x^{2}} \text { and } v=1\\ &\text { Hence } u^{\prime}=\frac{1}{2}\left(4-x^{2}\right)^{\frac{1}{2}-1}(-2 x)=\frac{-x}{\sqrt{4-x^{2}}}\\ &\Rightarrow \int_{-2}^{2} y d x=\left[\sqrt{4-x^{2}} \int 1 d x\right]_{-2}^{2}-\int_{-2}^{2}\left(\frac{-x}{\sqrt{4-x^{2}}}\right)(1 d x) d x\\ &\Rightarrow \int_{-2}^{2} y d x=\left[\sqrt{4-x^{2}}(x)\right]_{-2}^{2}-\int_{-2}^{2}\left(\frac{-x^{2}}{\sqrt{4-x^{2}}}\right) d x\\ &\Rightarrow \int_{-2}^{2} \mathrm{ydx}=\left(\sqrt{4-2^{2}}(2)\right)-\left(\sqrt{4-(-2)^{2}}(-2)\right)-\int_{-2}^{2}\left(\frac{4-\mathrm{x}^{2}-4}{\sqrt{4-\mathrm{x}^{2}}}\right) \mathrm{d} \mathrm{x} \end{aligned}$
$\\ \Rightarrow \int_{-2}^{2} y d x=-\int_{-2}^{2}\left(\frac{4-x^{2}}{\sqrt{4-x^{2}}}-\frac{4}{\sqrt{4-x^{2}}}\right) d x \\$
$ \Rightarrow \int_{-2}^{2} y d x=-\int_{-2}^{2} \sqrt{4-x^{2}} d x+\int_{-2}^{2} \frac{4}{\sqrt{4-x^{2}}} d x \\ \text { But } y=\sqrt{4-x^{2}} \\$
$ \Rightarrow \int_{-2}^{2} y d x=-\int_{-2}^{2} y d x+\int_{-2}^{2} \frac{4}{\sqrt{4-x^{2}}} d x \\$
$ \Rightarrow \int_{-2}^{2} y d x+\int_{-2}^{2} y d x=\int_{-2}^{2} \frac{4}{\sqrt{4-x^{2}}} d x \\$
$ \Rightarrow \int_{-2}^{2} y d x=2 \int_{-2}^{2} \frac{1}{\sqrt{2^{2}-x^{2}}} d x$
$\\\begin{aligned} &\text { We know that } \int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1} \frac{x}{a}\\ &\Rightarrow \int_{-2}^{2} \mathrm{y} \mathrm{dx}=2\left[\sin ^{-1} \frac{\mathrm{x}}{2}\right]_{-2}^{2}\\ &\Rightarrow \int_{-2}^{2} \mathrm{y} \mathrm{dx}=2\left(\sin ^{-1} \frac{2}{2}-\sin ^{-1} \frac{-2}{2}\right)\\ &\Rightarrow \int_{-2}^{2} \mathrm{y} \mathrm{dx}=2\left(\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right)\\ &\Rightarrow \int_{-2}^{2} \mathrm{y} \mathrm{dx}=2\left(\frac{\pi}{2}+\frac{\pi}{2}\right)\\ &\Rightarrow \int_{-2}^{2} \mathrm{y} \mathrm{dx}=2 \pi\\ &\text { Hence area is } 2 \pi \text { unit }^{2} \end{aligned}$
Question 9
Calculate the area under the curve $y=2\sqrt x$ included between the lines x = 0 and x = 1.
Answer:
$y=2 \sqrt{x}$ will be the parabolic curve of $y^2=4 x$ only in $1^{\text {st }}$ quadrant $\mathrm{x}=0$ is the equation of the $Y$-axis and $\mathrm{x}=1$ is a line parallel to Y -axis passing through $(1,0)$. Plot equations $y=2 \sqrt{x}$ and $\mathrm{X}=1$
So we have to integrate $y=2 \sqrt{x}$ from 0 to 1 let us find area under parabola
$
\Rightarrow y=2 \sqrt{x}
$
Integrate from 0 to 1
$
\begin{aligned}
& \Rightarrow \int_0^1 \mathrm{ydx}=\int_0^1 2 \mathrm{x}^{\frac{1}{2}} \mathrm{dx} \\
& \Rightarrow \int_0^1 \mathrm{ydx}=2\left[\frac{\mathrm{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_0^1 \\
& \Rightarrow \int_0^1 \mathrm{ydx}=2\left[\frac{\mathrm{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^1 \\
& \Rightarrow \int_0^1 \mathrm{ydx}=2 \frac{2}{3}\left[\mathrm{x}^{\frac{3}{2}}\right]_0^1 \\
& \Rightarrow \int_0^1 \mathrm{ydx}=\frac{4}{3}\left(1^{\frac{3}{2}}-0\right) \\
& \Rightarrow \int_0^2 \mathrm{ydx}=\frac{4}{3}
\end{aligned}
$
Therefore, the area found to be
$
\frac{4}{3} \text { unit }^2
$
Question 10
Answer:
The equation of the line is 2y = 5x + 7
Now, we need the point of the line, so we substitute x = 0 and then y = 0
$\\Put \: \: x=0 \Rightarrow 2 y=5(0)+7\\ \Rightarrow \mathrm{y}=\frac{7}{2}\\ Put\: \: y=0\\ \\\Rightarrow 2(0)=5 x+7 \\\Rightarrow 5 x=-7 \\\Rightarrow x=-\frac{7}{5}$
$\\Hence \left(0, \frac{7}{2}\right) and \left(-\frac{7}{5}, 0\right)$ are the required two points to draw the line $2 \mathrm{y}=5 \mathrm{x}+7$
The other two, x = 2 and x = 8, are straight lines parallel to the Y axis.
On plotting all the above lines.
$\\\text{We have to find area under} \ the \ \mathrm{y}=5 \mathrm{x}+7$ that is $\mathrm{y}=1 / 2(5 \mathrm{x}+7)$ from 2 to 8 $\Rightarrow y=1 / 2(5 x+7)$ Integrate from 2 to 8
$\Rightarrow \int_{2}^{8} \mathrm{ydx}=\frac{1}{2} \int_{2}^{8}(5 \mathrm{x}+7) \mathrm{dx}$
$\\\Rightarrow \int_{2}^{8} \mathrm{ydx}=\frac{1}{2}\left[\frac{5 \mathrm{x}^{2}}{2}+7 \mathrm{x}\right]_{2}^{8}$
$\\\Rightarrow \int_{2}^{8} \mathrm{ydx}=\frac{1}{2}\left[\left(\frac{5(8)^{2}}{2}+7(8)\right)-\left(\frac{5(2)^{2}}{2}+7(2)\right)\right]$
$\\\Rightarrow \int_{2}^{8} \mathrm{ydx}=\frac{1}{2}[(160+56)-(10+14)]$
$\\\begin{aligned} &\Rightarrow \int_{2}^{8} \mathrm{ydx}=\frac{1}{2}(192)\\ &\Rightarrow \int_{2}^{8} \mathrm{ydx}=96\\ &\text { Hence the area bounded by given lines is } 96 \text { unit }^{2} \end{aligned}$
Question 11
Answer:
$\\ \mathrm{y}=\sqrt{\mathrm{x}-1} \\$
Squaring both sides
$\Rightarrow y^{2}=x-1 \\ \mathrm{y}^{2}=\mathrm{x}-1\text{ is equation of a parabola} \\$
The above equation has no values for x less than 1; therefore, the parabola will be to the right of x = 1
Now observe that in $\mathrm{y}=\sqrt{\mathrm{x}-1} \mathrm{x} \geq 1 \: \: and\: \: \mathrm{y}$ has to positive because of square root hence $\mathrm{x} \ and\ \mathrm{y}$ both positive hence the parabola will be drawn only in $1^{\text {st }}$ quadrant
We have to plot the curve in [1,5], so just draw the parabolic curve from x=1 to x=5 in $1^{\text {st }}$ quadrant
x=1 and x=5 are lines parallel to the Y-axis
So we have to integrate $\mathrm{y}=\sqrt{\mathrm{x}-1}$ from 1 to 5
Let us find the area under the parabolic curve
$\Rightarrow y=\sqrt{x-1}$
Integrate from 1 to 5
$\Rightarrow \int_{1}^{5} y \mathrm{dx}=\int_{1}^{5}(\mathrm{x}-1)^{\frac{1}{2}} \mathrm{dx}\\ \Rightarrow \int_{1}^{5} y d x=\left[\frac{(x-1)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{1}^{5}\\ \Rightarrow \int_{1}^{5} \mathrm{ydx}=\left[\frac{(\mathrm{x}-1)^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{5} \\$
$\\\begin{array}{l} \Rightarrow \int_{1}^{5} y d x=\frac{2}{3}\left[(x-1)^{\frac{3}{2}}\right]_{1}^{5} \\ \Rightarrow \int_{1}^{5} y d x=\frac{2}{3}\left((5-1)^{\frac{3}{2}}-0\right) \\ \Rightarrow \int_{1}^{5} y d x=\frac{2}{3}\left(2^{2}\right)^{\frac{3}{2}} \\ \Rightarrow \int_{1}^{5} y d x=\frac{2}{3} \times 2^{3} \\ \Rightarrow \quad \int_{1}^{5} y d x=\frac{16}{3} \\ \text { Hence area bounded }=\frac{16}{3} \text { unit }^{2} \end{array} \\$
Question 12
Determine the area under the curve $y=\sqrt{a^2-x^2}$ included between the lines x = 0 and x = a
Answer:
$
y=\sqrt{a^2-x^2}
$
Squaring both sides
$
\begin{aligned}
& \Rightarrow y^2=a^2-x^2 \\
& \Rightarrow x^2+y^2=a^2
\end{aligned}
$
The above equation is of a circle having centre as $(0,0)$ and radius a .Now in $\mathrm{y}=\sqrt{\mathrm{a}^2-\mathrm{x}^2}-a \leq \mathrm{x} \leq \mathrm{a}$ and $\mathrm{y} \geq 0$ which means x and y both positive or x negative and y positive hence the curve $y=\sqrt{a^2-x^2}$ has to be above $X$-axis in $1^{\text {st }}$ and $2^{\text {nd }}$ quadrant
$x=0$ is equation of $Y$-axis and $x=a$ is a line parallel to $Y$-axis passing through $(a, 0)$
So we have to integrate $y=\sqrt{a^2-x^2}$ from 0 to a
So we have to integrate $y=\sqrt{a^2-x^2}$ from 0 to a
Let us find the area under the curve
$
y=\sqrt{a^2-x^2}
$
Integrate from 0 to a
$
\Rightarrow \int_0^a \mathrm{ydx}=\int_0^a \sqrt{\mathrm{a}^2-\mathrm{x}^2} \mathrm{dx}
$
Using the uv rule of integration where $u$ and $v$ are functions of
$
\int_a^b u v d x=\left[u \int v d x\right]_a^b-\int_a^b\left(u^{\prime} \int v d x\right) d x
$
Here $u=\sqrt{a^2-x^2}$ and $\mathrm{v}=1$
Hence
$
\begin{aligned}
& \mathrm{u}^{\prime}=\frac{1}{2}\left(\mathrm{a}^2-\mathrm{x}^2\right)^{\frac{1}{2}-1}(-2 \mathrm{x})=\frac{-\mathrm{x}}{\sqrt{\mathrm{a}^2-\mathrm{x}^2}} \\
& \Rightarrow \int_0^a y d x=\left[\sqrt{a^2-x^2} \int_0^a d x\right]_0^a-\int_0^a\left(\frac{-x}{\sqrt{a^2-x^2}} \int_0^a d x\right) d x \\
& \Rightarrow \int_0^a y d x=\left[\sqrt{a^2-x^2}(x)\right]_0^a-\int_0^a\left(\frac{-x^2}{\sqrt{a^2-x^2}}\right) \mathrm{d} x \\
& \Rightarrow \int_0^a y d x=\left(\sqrt{a^2-a^2}(a)\right)-\left(\sqrt{a^2-0^2}(0)\right)-\int_0^a\left(\frac{a^2-x^2-a^2}{\sqrt{a^2-x^2}}\right) \mathrm{d} x \\
& \Rightarrow \int_0^a y d x=-\int_0^a\left(\frac{a^2-x^2}{\sqrt{a^2-x^2}}-\frac{a^2}{\sqrt{a^2-x^2}}\right) d x \\
& \Rightarrow \int_0^a y d x=-\int_0^a \sqrt{a^2-x^2} d x+\int_0^a \frac{a^2}{\sqrt{a^2-x^2}} d x \\
& \text { But } y=\sqrt{a^2-x^2}
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow \int_0^a y d x=-\int_0^a y d x+\int_0^a \frac{a^2}{\sqrt{a^2-x^2}} d x \\
& \Rightarrow \int_0^a y d x+\int_0^a y d x=\int_0^a \frac{a^2}{\sqrt{a^2-x^2}} d x \\
& \Rightarrow 2 \int_0^a y d x=a^2 \int_0^a \frac{1}{\sqrt{a^2-x^2}} d x \\
& \Rightarrow \int_0^a y d x=\frac{a^2}{2} \int_0^a \frac{1}{\sqrt{a^2-x^2}} d x
\end{aligned}
$
We know that $\int_0^a \frac{1}{\sqrt{a^2-x^2}} d x=\sin ^{-1} \frac{x}{a}$
$
\begin{aligned}
& \Rightarrow \int_0^a y d x=\frac{a^2}{2}\left[\sin ^{-1} \frac{x}{a}\right]_0^a \\
& \Rightarrow \int_0^a y d x=\frac{a^2}{2}\left(\sin ^{-1} \frac{a}{a}-\sin ^{-1} \frac{0}{2}\right) \\
& \Rightarrow \int_0^a y d x=\frac{a^2}{2}\left(\frac{\pi}{2}-0\right) \\
& \Rightarrow \int_0^a y d x=\frac{\pi a^2}{4}
\end{aligned}
$
Hence area bounded $=\frac{\pi a^2}{4}$ unit ${ }^2$
Question 13
Find the area of the region $y=\sqrt x$ bounded by and y = x.
Answer:
$\begin{aligned} &y=\sqrt{x}\\ &\text { squaring both sides }\\ &\Rightarrow y^{2}=x \end{aligned} \\$
This is a parabola, with no negative values of x, it lies on the right of the Y axis, passing through the origin.
Now $y=\sqrt{x}$ means y and x both has to be positive hence both lie in $1^{\text {st }}$ quadrant hence $y=\sqrt{x}$ will be part of $\mathrm{y}^{2}=\mathrm{x}$ which is lying only in $1^{\text {st }}$ quadrant
And y=x is a straight line passing through the origin
We have to find area between $y=\sqrt{x}$ and y=x shown below
To find the point of interaction, solve two equations simultaneously.
$\\Put y=x in y^{2}=x \\ \Rightarrow x^{2}=x \\ \Rightarrow x=1 \\$
Put x=1 in y=x we get y=1
The point of interaction is (1, 1)
Area between the parabolic curve and line = area under parabolic curve – area under line $\ldots (1) \\$
For the area under the parabolic curve
$\int Y = \sqrt x \\$
Integrating from 0 to 1
$\begin{aligned} &\Rightarrow \int_{0}^{1} \mathrm{ydx}=\int_{0}^{1} \mathrm{x}^{\frac{1}{2}} \mathrm{~d} \mathrm{x}\\ &\Rightarrow \int_{0}^{1} \mathrm{ydx}=\left[\frac{\mathrm{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{1}\\ &\Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=\left[\frac{\mathrm{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1}\\ &\Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=\frac{2}{3}\left[1^{\frac{3}{2}}-0\right]\\ &\Rightarrow \int_{0}^{1} \mathrm{ydx}=\frac{2}{3} \end{aligned} \\$
For the area under the straight line y = x
On integrating from 0 to 1
$\Rightarrow \int_{0}^{1} \mathrm{ydx}=\int_{0}^{1} \mathrm{xdx} \\ \Rightarrow \int_{0}^{1} y d x=\left[\frac{x^{2}}{2}\right]_{0}^{1} \\ \Rightarrow \int_{0}^{1} \mathrm{ydx}=\left(\frac{1^{2}}{2}-0\right) \\ \Rightarrow \int_{0}^{1} y d x=\frac{1}{2} \\$
Using (i)
$\Rightarrow \text{area between parabolic } = \frac{2}{3}-\frac{1}{2}= \frac{1}{6} unit ^{2} \\$
Hence area bounded is $\frac{1}{6} unit ^{2} \\ \\$
Question 14
Find the area enclosed by the curve $y = -x^2$ and the straight line x + y + 2 = 0.
Answer:
$\\Y = -x^{2 }\\ X^{2 }= -y \\$
This equation depicts a parabola defining no positive values of y therefore it lies below X axis and passes through origin \\
X + y + 2 = 0 depicts a straight line
For the point of interaction, both of the equations are solved simultaneously.
$\begin{array}{l} \text { Put } y=-(x+2) \text { in } x^{2}=-y \\ \Rightarrow x^{2}=-(-(x+2)) \\ \Rightarrow x^{2}=x+2 \\ \Rightarrow x^{2}-x-2=0 \\ \Rightarrow(x+1)(x-2)=0 \\ \Rightarrow x^{2}-2 x+x-2=0 \\ \Rightarrow x(x-2)+1(x-2)=0 \\ \Rightarrow x=-1 \text { and } x=2 \\\end{array} \\$
$\begin{array}{l} \Rightarrow y=-1 \\ \text { Put } x=2 \text { in } x^{2}=-y \\ \Rightarrow 2^{2}=-y \\ \Rightarrow y=-4 \end{array} \\$
Therefore, the points of interaction are (-1, -1) and (2, -4)
The figure below shows the area to be calculated
Area between the line and parabola = Area under line – area under parabola $\ldots (1) \\$
To find the area under the line, integrate it from -1 to 2
Y = -(x + 2)
$\begin{array}{l} \Rightarrow \int_{-1}^{2} \mathrm{y} \mathrm{dx}=\int_{-1}^{2}-(\mathrm{x}+2) \mathrm{d} \mathrm{x} \\ \Rightarrow \int_{-1}^{2} \mathrm{ydx}=-\left[\frac{\mathrm{x}^{2}}{2}+2 \mathrm{x}\right]_{-1}^{2} \\ \Rightarrow \int_{-1}^{2} \mathrm{y} \mathrm{dx}=-\left[\left(\frac{2^{2}}{2}+2(2)\right)-\left(\frac{-1^{2}}{2}+2(-1)\right)\right] \\ \Rightarrow \int_{-1}^{2} \mathrm{y} \mathrm{d} \mathrm{x}=-\left[6-\left(\frac{1}{2}-2\right)\right] \\ \Rightarrow \int_{-1}^{2} \mathrm{y} \mathrm{d} \mathrm{x}=-\frac{15}{2} \end{array} \\$
For the area under a parabola, integrate it from -1 to 2
$Y = -x^{2}\\$
$\begin{aligned} &\text { Integrate from }-1 \text { to } 2\\ &\Rightarrow \int_{-1}^{2} y d x=\int_{-1}^{2}-x^{2} d x\\ &\Rightarrow \int_{-1}^{2} \mathrm{ydx}=-\left[\frac{\mathrm{x}^{3}}{3}\right]_{-1}^{2}\\ &\Rightarrow \int_{-1}^{2} \mathrm{ydx}=-\left(\frac{2^{3}}{3}-\frac{(-1)^{3}}{3}\right)\\ &\Rightarrow \int_{-1}^{2} y d x=-\left(\frac{8}{3}+\frac{1}{3}\right)\\ &\Rightarrow \int_{-1}^{2} y d x=-3 \end{aligned} \\$
Substituting both values in (1)
area enclosed by line and parabola $=-\frac{15}{2}-(-3)=-\frac{9}{2} \text { unit }^{2}$
The negative sign depicts the area below the X-axis
Hence the area enclosed is $\frac{9}{2} \text { unit }^{2}$
Question 15
Find the area bounded by the curve $y=\sqrt x$, x = 2y + 3 in the first quadrant and x-axis.
Answer:
$\begin{aligned} &y=\sqrt{x}\\ &\text { Squaring both sides }\\ &\Rightarrow y^{2}=x \end{aligned} \\$
The above equation depicts a parabola which does not define negative values for x and therefore, it lies to the right of the Y-axis and passes through the origin.
For plotting $y = \sqrt x$, both the values of x and y have to be positive; hence, this graph lies in the 1st quadrant only.
The point cannot be negative because the square root symbol itself depicts the positive root.
The equation x = 2y + 3 is a straight line.
To find the point of interaction, solve the two equations simultaneously.
$\\ \text { Put } x=2 y+3 \text { in } y^{2}=x \\ \Rightarrow y^{2}=2 y+3 \\ \Rightarrow y^{2}-2 y-3=0 \\ \Rightarrow y^{2}-3 y+y-3=0 \\ \Rightarrow y(y-3)+1(y-3)=0 \\ \Rightarrow(y+1)(y-3)=0 \\ \Rightarrow y=-1 \text { and } y=3 \\ \text { Put } y=3 \text { in } y^{2}=x \\ \Rightarrow x=3^{2} \\ \Rightarrow x=9 \\$
Y = -1 won’t make a difference since we are looking for the 1st quadrant only.
Therefore, the points of interaction are (9, 3)
Below is the figure showing the area to be calculated
The point of the intersection of a straight line on X axis can be calculated by substituting y = 0 in the equation of the line
$\int X = 2(0) + 3 => x = 3$
The area enclosed = area under $y = \sqrt x$ – area under the straight line $\ldots (1) \\$
For the area under $y = \sqrt x$ integrate the equation from 0 to 9
$\begin{aligned} &\Rightarrow \int_{0}^{9} y d x=\int_{0}^{9} x^{\frac{1}{2}} d x\\ &\Rightarrow \int_{0}^{9} y d x=\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{9}\\ &\Rightarrow \int_{0}^{9} \mathrm{ydx}=\left[\frac{\mathrm{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{9}\\ &\Rightarrow \int_{0}^{9} y d x=\frac{2}{3}\left[9^{\frac{3}{2}}-0\right]\\ &\Rightarrow \int_{0}^{9} y d x=\frac{2}{3}\left[\left(3^{2}\right)^{\frac{3}{2}}\right]\\ &\Rightarrow \int_{0}^{9} y d x=\frac{2}{3} \times 3^{3} \end{aligned} \\$
For the area under a straight line
Y = (x-3) / 2
Integrating the equation from 3 to 9
$\\ \Rightarrow \int_{3}^{9} y d x=\frac{1}{2} \int_{3}^{9}(x-3) d x \\ \Rightarrow \int_{3}^{9} y d x=\frac{1}{2}\left[\frac{x^{2}}{2}-3 x\right]_{3}^{9} \\$
$ \Rightarrow \int_{3}^{9} y d x=\frac{1}{2}\left(\left(\frac{9^{2}}{2}-3(9)\right)-\left(\frac{3^{2}}{2}-3(3)\right)\right) \\$
$ \Rightarrow \int_{3}^{9} y d x=\frac{1}{2}\left(\left(\frac{81}{2}-27\right)-\left(\frac{9}{2}-9\right)\right) \\$
$ \Rightarrow \int_{0}^{9} y d x=\frac{1}{2} \times 18$
$\Rightarrow \int_{0}^{9} \mathrm{ydx}=9$
Using (i)
$\Rightarrow$ area bounded =18-9=9 $unit ^{2}$
Hence area bounded by the curve and a straight line is 9 $unit ^{2}$
Question 16
Find the area of the region bounded by the curve $y^2 = 2x$ and $x^2 + y^2 = 4x$.
Answer:
$y^2=2 x$ depicts a parabola having no negative values for x and lying to the right of the Y axis while passing through the origin. The other equation of $x^2+y^2=4 x$ depicts a circle.
The general equation of a circle is given by $x^2+y^2+2 g x+2 f y+c=0$
Centre of circle is $(-g,-f)$ and radius is $\sqrt{g^2+f^2-c}$
$
\text { In } x^2+y^2-4 x=0,2 g=-4 \Rightarrow g=-2 \text { and } f=c=0
$
Hence center is $(-(-2), 0)$ that is $(2,0)$ and radius is $\sqrt{(-2)^2+0^2-0}$ which is 2
For the point of interaction, solve the two equations simultaneously.
$
\begin{aligned}
& \text { Put } y^2=2 x \text { in } x^2+y^2=4 x \\
& \Rightarrow x^2+2 x=4 x \\
& \Rightarrow x^2-2 x=0 \\
& \Rightarrow x(x-2)=0 \\
& \Rightarrow x=0 \text { and } x=2 \\
& \text { Put } x=2 \text { in } y^2=2 x \\
& \Rightarrow y= \pm 2
\end{aligned}
$
Therefore, the point of interaction are $(0,0),(2,2)$, and $(2,-2)$
Below is the diagram of the area to be calculated.
On integrating, we will get an area for the 1st quadrant only, but since it is symmetrical, we can multiply it by.
Area of shaded region $=$ area under the circle - area under the parabola $\ldots$ (1)To findg the area under circle
$
\begin{aligned}
& x^2+y^2=4 x \\
& \Rightarrow y^2=4 x-x^2 \\
& \Rightarrow y=\sqrt{4 x-x^2} \\
& \Rightarrow y=\sqrt{-\left(-4 x+x^2+4-4\right)} \\
& \Rightarrow y=\sqrt{-\left(\left(x^2-4 x+4\right)-2^2\right)} \\
& \Rightarrow y=\sqrt{-\left((x-2)^2-2^2\right)} \\
& \Rightarrow y=\sqrt{2^2-(x-2)^2}
\end{aligned}
$
Integrate the above equation from 0 to 2
$
\Rightarrow \int_0^2 y d x=\int_0^2 \sqrt{2^2-(x-2)^2} d x
$
$
\text { Using } \int \sqrt{a^2-x^2}=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}
$
$
\Rightarrow \int_0^2 y d x=\left[\frac{x-2}{2} \sqrt{2^2-(x-2)^2}+\frac{2^2}{2} \sin ^{-1} \frac{x-2}{2}\right]_0^2
$
$
\Rightarrow \int_0^2 \mathrm{ydx}=\left[0-\left(\frac{0-2}{2} \sqrt{2^2-(0-2)^2}+\frac{2^2}{2} \sin ^{-1} \frac{0-2}{2}\right)\right]
$
$
\Rightarrow \int_0^2 y d x=\left[-\left(0+2 \sin ^{-1}(-1)\right)\right]
$
$
\Rightarrow \int_0^2 y d x=\pi
$
For the area under the parabola,
$
\begin{aligned}
& \Rightarrow y^2=2 x \\
& \Rightarrow y=\sqrt{2} \sqrt{x}
\end{aligned}
$
Integrate the above equation from 0 to 2
$
\begin{aligned}
& \Rightarrow \int_0^2 y d x=\sqrt{2} \int_0^2 x^{\frac{1}{2}} d x \\
& \Rightarrow \int_0^2 y d x=\sqrt{2}\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_0^2 \\
& \Rightarrow \int_0^2 y d x=2^{\frac{1}{2}}\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^2 \\
& \Rightarrow \int_0^2 \mathrm{ydx}=2^{\frac{1}{2}} \times \frac{2}{3}\left[\mathrm{x}^{\frac{3}{2}}\right]_0^2 \\
& \Rightarrow \int_0^2 \mathrm{ydx}=\frac{2^{\frac{1}{2}+1}}{3}\left(2^{\frac{3}{2}}-0\right) \\
& \Rightarrow \int_0^2 \mathrm{ydx}=\frac{2^{\frac{3}{2}+\frac{3}{2}}}{3} \\
& \Rightarrow \int_0^2 \mathrm{ydx}=\frac{2^3}{3} \\
& \Rightarrow \int_0^2 \mathrm{ydx}=\frac{8}{3}
\end{aligned}
$
Using (i)
$\Rightarrow$ area of shaded in $1^{\text {st }}$ quadrant $=\pi-\frac{8}{3}$ unit $^2$
After multiplying it by 2
The area required of shaded region $=2\left(\pi-\frac{8}{3}\right)$ unit $^2$
Question 17
Find the area bounded by the curve y = six between x = 0 and x = $2 \pi$
Answer:
Below is the graph with the shaded region whose area has to be calculated
From the plot, it is seen that the area from 0 to $\pi$ is positive, whereas the area from $\pi$ to 2$\pi$ is negative.
But both areas are equal in magnitude with opposite signs.
If we integrate them, the areas will cancel each other, and the answer will be 0.
Therefore, we integrate the two areas separately and split the limits in two ways –
1) Find the area under six from 0 to $\pi$ and multiply by 2
2)Split the limit 0 to 2 $\pi$ into 0 to $\pi$ and $\pi$ to 2 $\pi$
Here, we will solve this by splitting the limits
$Y = six \\$
$\begin{aligned} &\text { Integrating from } 0 \text { to } 2 \pi\\ &\Rightarrow \int_{0}^{2 \pi} \mathrm{ydx}=\int_{0}^{2 \pi} \sin \mathrm{xdx}\\ &\Rightarrow \int_{0}^{2 \pi} \mathrm{ydx}=\int_{0}^{\pi} \sin \mathrm{x} \mathrm{d} \mathrm{x}+\int_{\pi}^{2 \pi} \sin \mathrm{xdx}\\ &\text { Because } \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\int_{\mathrm{a}}^{\mathrm{c}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}+\int_{\mathrm{c}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x} \text { where } c \in(\mathrm{a}, \mathrm{b}) \end{aligned} \\$
Now the limits become negative
$\begin{aligned} &\begin{array}{l} \text { Hence for } \pi \text { to } 2 \pi \sin x \text { will become }-\sin x \end{array}\\ &\Rightarrow \int_{0}^{2 \pi} \mathrm{y} \mathrm{dx}=\int_{0}^{\pi} \sin \mathrm{x} \mathrm{d} \mathrm{x}+\int_{\pi}^{2 \pi}-\sin \mathrm{x} \mathrm{d} \mathrm{x}\\ &\Rightarrow \int_{0}^{2 \pi} \mathrm{ydx}=\int_{0}^{\pi} \sin \mathrm{xdx}-\int_{\pi}^{2 \pi} \sin \mathrm{xdx}\\ &\Rightarrow \int_{0}^{2 \pi} \mathrm{ydx}=[-\cos \mathrm{x}]_{0}^{\pi}-[-\cos \mathrm{x}]_{\pi}^{2 \pi}\\ &\Rightarrow \int_{0}^{2 \pi} \mathrm{ydx}=(-(\cos \pi-\cos 0))-(-(\cos 2 \pi-\cos \pi))\\ &\Rightarrow \int_{0}^{2 \pi} \mathrm{ydx}=(-((-1)-1))-(-(1-(-1))) \end{aligned} \\$
$\Rightarrow \int_{0}^{2 \pi} \mathrm{ydx}=4 $
Hence area bounded by six from 0 to 2 $\pi$ $\text{ is }4 unit $^{2}$ \\$
Question 18
Answer:
The vertices of the triangle are given as (-1, 1), (0, 5) and (3, 2)
The three vertices are given alphabets as P, Q and R, respectively.
$\Rightarrow$ Equation of PQ is $y-1=\frac{5-1}{0+1}(x+1)$thereforey $=4 x+5 \Rightarrow$ Equation of $Q R$ is $y-5=\frac{2-5}{3-0}(x-0) \therefore y=-x+5$
$\Rightarrow$ Equation of RP is $y-2=\frac{1-2}{-1-3}(x-3) \therefore 4 y=x+5$
Area of the region bounded by the curve $y=f(x)$, the x-axis and the ordinates $\mathrm{x}=\mathrm{a}$ and $\mathrm{x}=\mathrm{b}$, where $\mathrm{f}(\mathrm{x})$ is a continuous function defined on [a, b], is given by $A=\int_a^b f(x) d x$ or $\int_a^b y d x$
Required area
$
\begin{aligned}
& =\int_{-1}^0(4 x+5) d x+\int_0^3(-x+5) d x-\int_{-1}^3\left(\frac{x}{4}+\frac{5}{4}\right) d x \\
& =\left[2 x^2+5 x\right]_{-1}^0+\left[-\frac{x^2}{2}+5 x\right]_0^3-\left[\frac{x^2}{8}+\frac{5 x}{4}\right]_{-1}^3 \\
& =((0+0)-(2-5))+\left(\left(-\frac{9}{2}+15\right)-(0+0)\right)-\left(\left(\frac{9}{8}+\frac{15}{4}\right)-\left(\frac{1}{8}-\frac{5}{4}\right)\right) \\
& =3+\frac{21}{2}-\frac{39}{8}-\frac{9}{8} \\
& =\frac{15}{2} \text { sq.units }
\end{aligned}
$
Question 19
Answer:
$\text { The region }\left\{(x, y): y^{2} \leq 6 a x \text { and } x^{2}+y^{2} \leq 16 a^{2}\right\}$
By solving the equations:
$
y^2 \leq 6 a x \quad \text { and } \quad x^2+y^2 \leq 16 a^2
$
Through substituting for $\mathrm{y}^2$
$
\begin{aligned}
& \Rightarrow x^2+6 a x=16 a^2 \\
& \Rightarrow(x-2 a)(x+8 a)=0 \\
& \therefore x=2 a
\end{aligned}
$
[ as $x=-8 a$ is not possible ]
Area of the region bounded by the curve $y=f(x)$, the $x$-axis and the ordinates $x=a$ and $x=b$, where $f(x)$ is a continuous function defined on $[a, b]$, is given by $A=\int_a^b f(x) d x$ or $\int_a^b y d x$.
[By the symmetry of the image w.r.t $x$ axis]
$
\left[\int \sqrt{a^2-x^2} d x=\frac{x \sqrt{a^2-x^2}}{2}+\frac{a^2}{2} \sin ^{-1}\left(\frac{x}{a}\right)\right]
$
Required area $=2\left[\int_0^{2 a} \sqrt{6 a x} d x+\int_{2 a}^{4 a} \sqrt{(4 a)^2-x^2} d x\right]$
$
\begin{aligned}
& =2\left[\sqrt{6 a}\left[\frac{2 x^{\frac{3}{2}}}{3}\right]_0^{2 a}+\left[\frac{x \sqrt{(4 a)^2-x^2}}{2}+\frac{(4 a)^2}{2} \sin ^{-1}\left(\frac{x}{4 a}\right)\right]_{2 a}^{4 a}\right] \\
& =2\left(\frac{8}{3} \sqrt{3} a^2+4 \pi a^2-2 \sqrt{3} a^2-\frac{4 a^2 \pi}{3}\right) \\
& =\frac{4}{3} a^2[\sqrt{3}+4 \pi] \text { sq.units }
\end{aligned}
$
Question 20
Compute the area bounded by the lines x + 2y = 2, y – x = 1 and 2x + y = 7.
Answer:
The lines given are x + 2y = 2
$x = 2 - 2y \ldots .(1) \\$
y – x = 1
$x = y -1 \ldots .(2) \\$
and $2x + y = 7 \ldots .(3) \\$
Equate the values of x from 1 and 2 to get,
$
\begin{aligned}
& 2-2 y=y-1 \\
& 2+1=y+2 y \\
& 3=3 y \\
& y=1
\end{aligned}
$
Put the value of $y$ in (2) to get
$
\begin{aligned}
& x=1-1 \\
& =0
\end{aligned}
$
So, the intersection point is $(0,1)$
On solving the equations, the point of interaction is found to be $(0,1),(2,3)$ and $(4,-1)$The area of the region bounded by the curve $y=f(x)$, the $x$-axis and the ordinates $x=a$ and $x=b$, where $f(x)$ is a continuous function defined on [a, b], is given by $A=\int_a^b f(x) d x$ or $\int_a^b y d x$
Required area
$
\begin{aligned}
& =\int_0^2\left(\mathrm{x}-1-\frac{2-\mathrm{x}}{2}\right) \mathrm{dx}+\int_2^4\left(7-2 \mathrm{x}-\frac{2-\mathrm{x}}{2}\right) \mathrm{dx} \\
& =\int_0^2\left(\frac{3 x}{2}\right) d x+\int_2^4\left(6-\frac{3 x}{2}\right) d x \\
& =\left[\frac{3 x^2}{4}\right]_0^2+\left[6 x-\frac{3 x^2}{4}\right]_2^4 \\
& =3+(24-12)-(12-3)=6 \text { sq.units }
\end{aligned}
$
Question 21
Find the area bounded by the lines y = 4x + 5, y = 5 - x and 4y = x + 5
Answer:
The lines mentioned are y = 4x + 5, y = 5 - x and 4y = x + 5
Below is the figure,
By solving these equations,
$
\begin{aligned}
& y=4 x+5 \ldots \ldots \\
& y=5-x \ldots \ldots(2) \\
& 4 y=x+5 \ldots
\end{aligned}
$
From (1) and (2)
$
\begin{aligned}
& 4 x+5=5-x \Rightarrow x=0 \\
& \therefore y=5-x=5
\end{aligned}
$
From (2) and (3)
$
\begin{aligned}
& 4(5-x)=x+5 \\
& \Rightarrow x=3 ; \therefore y=5-x=2 \\
& \text { From }(1) \text { and }(3) \\
& 4(4 x+5)=x+5 \\
& \Rightarrow x=-1 ; \therefore y=4 x+5=1
\end{aligned}
$
The point of interaction have found to be $(0,5),(3,2)$ and $(-1,1)$
Area of the region bounded by the curve $y=f(x)$, the $x$-axis and the ordinates $x=a$ and $x=b$, where $f(x)$ is a continuous function defined on $[\mathrm{a}, \mathrm{b}]$, is given by $A=\int_a^b f(x) d x$ or $\int_a^b y d x$
Required area
$
\begin{aligned}
& =\int_{-1}^0(4 x+5) d x+\int_0^3(-x+5) d x-\int_{-1}^3\left(\frac{x}{4}+\frac{5}{4}\right) d x \\
& =\left[2 x^2+5 x\right]_{-1}^0+\left[-\frac{x^2}{2}+5 x\right]_0^3-\left[\frac{x^2}{8}+\frac{5 x}{4}\right]_{-1}^3 \\
& =((0+0)-(2-5))+\left(\left(-\frac{9}{2}+15\right)-(0+0)\right)-\left(\left(\frac{9}{8}+\frac{15}{4}\right)-\left(\frac{1}{8}-\frac{5}{4}\right)\right) \\
& =3+\frac{21}{2}-\frac{39}{8}-\frac{9}{8} \\
& =\frac{15}{2} \text { sq. units }
\end{aligned}
$
Question 22
Find the area bounded by the curve y = 2cos x and the x-axis from x = 0 to x = 2 $\pi$
Answer:
The curve given is y = 2cosx and xx-axis from x = 0 to x = 2 $\pi$
Below is the figure,
The area of the region bounded by the curve y=f(x), the x-axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a, b], is given by $A=\int_{a}^{b} f(x) d x or \int_{a}^{b} y d x$
From the figure;
Required area $=\int_{0}^{2 \pi}|2 \cos x| \mathrm{d} \mathrm{x}=4 \int_{0}^{\frac{\pi}{2}}(2 \cos \mathrm{x}) \mathrm{dx}$
$\\=8[\sin \mathrm{x}]_{0}^{\frac{\pi}{2}}\\ =8 sq.units \\$
Question 23
Answer:
Given;
Curve $y = 1 + \vert x +1 \vert , x = -3, x = 3, y = 0$
$|x+1|=\left\{\begin{array}{l}-x-1, x<-1 \\ x+1, x \geq-1\end{array}\right. \therefore y=1+|x+1| =\left\{\begin{array}{c}-x, x<-1 \\ x+2, x \geq-1\end{array}\right.$
AreThe area of the region bounded by the curve y=f(x), the x-axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a, b], is given by $A=\int_{a}^{b} f(x) d x or \int_{a}^{b} y d x \\$
Required area
$=\int_{-3}^{-1}(-\mathrm{x}) \mathrm{dx}+\int_{-1}^{3}(\mathrm{x}+2) \mathrm{d} \mathrm{x}\\ =-\left[\frac{\mathrm{x}^{2}}{2}\right]_{-3}^{-1}+\left[\frac{\mathrm{x}^{2}}{2}+2 \mathrm{x}\right]_{-1}^{3}\\ =-\left(\frac{1}{2}-\frac{9}{2}\right)+\left(\frac{9}{2}+6-\frac{1}{2}+2\right)\\ =4+12=16 sq.units \\$
Question 24
The area of the region bounded by the y-axis, y = cos x and y = sin x, $0 \leq x \leq\frac{\pi}{2}$ is.
A. $\sqrt{2}$ sq units
B. ($\sqrt{2}$ + 1) sq units
C. ($\sqrt{2}$ – 1) sq unit
D. (2$\sqrt{2}$ – 1) sq unit
Answer:
$y-axis, y = cosx$ and $y = sinx, 0 \leq x \leq \pi /2 \\$
$\begin{aligned} &\Rightarrow \sin x=\cos x\\ &\therefore x=\frac{\pi}{4}\\ &\text { Required area }\\ &=\int_{0}^{\frac{\pi}{4}}(\cos x-\sin x) d x\\ &=[\sin \mathrm{x}+\cos \mathrm{x}]_{0}^{\frac{\pi}{4}}\\ &=\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-0-1\right)\\ &=(\sqrt{2}-1) \text { sq.units } \end{aligned}$
Question 25
The area of the region bounded by the curve $x^2 = 4y$ and the straight line x = 4y – 2 is.
A. $\frac{3}{8}$ sq units
B.$\frac{5}{8}$ sq units
C. $\frac{7}{8}$ sq units
D. $\frac{9}{8}$ sq units
Answer:
The questions gives equation for curve $x^2 = 4y$ and straight line x = 4y – 2
Below is the figure,
By substituting for $4 \mathrm{y} ; \Rightarrow x=x^2-2$
$
\begin{aligned}
& \Rightarrow x^2-x-2=0 \\
& \Rightarrow(x+1)(x-2)=0 \\
& \therefore x=-1,2
\end{aligned}
$
Required area
$
\begin{aligned}
& =\int_{-1}^2\left(\frac{x+2}{4}\right) d x-\int_{-1}^2 \frac{x^2}{4}=\frac{1}{4}\left[\frac{x^2}{2}+2 x\right]_{-1}^2-\frac{1}{4}\left[\frac{x^3}{3}\right]_{-1}^2 \\
& =\frac{1}{4}\left(\frac{4}{2}+4-\frac{1}{2}+2-\frac{8}{3}-\frac{1}{3}\right) \\
& =\frac{9}{8} \text { sq. units }
\end{aligned}
$
Question 26
The area of the region bounded by the curve $y=\sqrt{16-x^2}$ and x-the axis is.
A. 8π sq units
B. 20π sq units
C. 16π sq units
D. 256π sq units
Answer:
$\text { The curve } y=\sqrt{16-x^{2}} \text { and } x \text { -axis; } y=0$
$
\begin{aligned}
& \Rightarrow \sqrt{16-x^2}=0 \\
& \therefore x= \pm 4
\end{aligned}
$
Required area
$
\begin{aligned}
& =\int_{-4}^4\left(\sqrt{16-x^2}\right) \mathrm{dx} \\
& {\left[\int \sqrt{\mathrm{a}^2-\mathrm{x}^2} \mathrm{dx}=\frac{\mathrm{x} \sqrt{\mathrm{a}^2-x^2}}{2}+\frac{\mathrm{a}^2}{2} \sin ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)\right]} \\
& =2 \int_0^4\left(\sqrt{4^2-x^2}\right) d x \\
& =2\left[\frac{\mathrm{x} \sqrt{4^2-\mathrm{x}^2}}{2}+\frac{4^2}{2} \sin ^{-1}\left(\frac{\mathrm{x}}{4}\right)\right]_0^4 \\
& =2\left(0+\frac{8 \pi}{2}-0-0\right) \\
& =8 \pi \text { sq.units }
\end{aligned}
$
Question 27
Area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle $x^2 + y^2 = 32$ is
A. 16π sq units
B. 4π sq units
C. 32π sq units
D. 24 sq units
Answer:
B)
The x-axis, the line y = x and the circle $x^2 + y^2 = 32$.
By substitution; $\Rightarrow x^2+x^2=32$
$
\therefore x=4
$
Radius of the circle $=\sqrt{32}=4 \sqrt{2}$
Required area
$
\begin{aligned}
& =\int_0^4 \mathrm{xdx}+\int_4^{4 \sqrt{2}}\left(\sqrt{32-\mathrm{x}^2}\right) \mathrm{dx} \\
& {\left[\int \sqrt{\mathrm{a}^2-\mathrm{x}^2} \mathrm{dx}=\frac{\mathrm{x} \sqrt{\mathrm{a}^2-\mathrm{x}^2}}{2}+\frac{\mathrm{a}^2}{2} \sin ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)\right]} \\
& =\int_0^4 x d x+\int_4^{4 \sqrt{2}}\left(\sqrt{(4 \sqrt{2})^2-x^2}\right) d x \\
& =\left[\frac{\mathrm{x}^2}{2}\right]_0^4+\left[\frac{\mathrm{x} \sqrt{(4 \sqrt{2})^2-\mathrm{x}^2}}{2}+\frac{(4 \sqrt{2})^2}{2} \sin ^{-1}\left(\frac{\mathrm{x}}{4 \sqrt{2}}\right)\right]_4^{4 \sqrt{2}} \\
& =\left(8-0+\left(\frac{32}{2} \times \frac{\pi}{2}\right)-\frac{16}{2}-\left(\frac{32}{2} \times \frac{\pi}{4}\right)\right) \\
& =(8+8 \pi-8-4 \pi) \\
& =4 \pi \text { sq.units }
\end{aligned}
$
Question 28
The area of the region bounded by the curve y = cos x between x = 0 and x = π is
A. 2 sq units
B. 4 sq units
C. 3 sq units
D. 1 sq units
Answer:
A)
The question mentions the curve y = cos x and x = 0 and x = π
Required area
$
\begin{aligned}
& =\int_0^\pi|\cos x| d x \\
& =2 \int_0^{\frac{\pi}{2}} \cos x d x \\
& =2[\sin x]_0^{\frac{\pi}{2}} \\
& =2(1-0) \\
& =2 \text { sq. units }
\end{aligned}
$
Question 29
The area of the region bounded by parabola $y^{2} = x$ and the straight line 2y = x is
A. $\frac{4}{3}$ sq units
B. 1 sq units
C. $\frac{2}{3}$ sq units
D. $\frac{1}{3}$ sq units
Answer:
A)
The questions gives parabola $y^2 = x$ and a straight line 2y = x
By substituting for x
$
\begin{aligned}
& \Rightarrow y^2=2 y \\
& \Rightarrow y(y-2)=0 \\
& \therefore y=0,2 \\
& \therefore x=2 \\
& y=0,4
\end{aligned}
$
Required area
$
\begin{aligned}
& =\int_0^4\left[\sqrt{\mathrm{x}}-\frac{\mathrm{x}}{2}\right] \mathrm{dx} \\
& =\left[\frac{2 \mathrm{x} \frac{3}{2}}{3}-\frac{\mathrm{x}^2}{4}\right]_0^4 \\
& =\left(\frac{16}{3}-4-0+0\right) \\
& =\frac{4}{3} \text { sq. units }
\end{aligned}
$
Question 30
The area of the region bounded by the curve y = sin x between the ordinates x = 0, x = π/2 and the x-axis is
A. 2 sq units
B. 4 sq units
C. 3 sq units
D. 1 sq units
Answer:
D)
The questions mentioned = sin x between the ordinates x = 0, x = π/2 and x x-axis.
$\\ =\int_{0}^{\frac{\pi}{2}}|\cos x| d x \\ =\int_{0}^{\frac{\pi}{2}} \sin x d x \\ =[-\cos x]_{0}^{\frac{\pi}{2}} \\ =0-(-1) \\ =1 \text { sq.units }$
Question 31
The area of the region bounded by the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$ is
A. 20π sq units
B. 20π2 sq units
C. 16π2 sq units
D. 25π sq units
Answer:
A)
The ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$
An ellipse has symmetry with the x-axis and the y-axis
The required area can be calculated as
$\\ =4 \int_{0}^{5} \frac{4}{5}\left(\sqrt{25-x^{2}}\right) d x \\ {\left[\int \sqrt{a^{2}-x^{2}} d x=\frac{x \sqrt{a^{2}-x^{2}}}{2}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)\right]} \\ =\frac{16}{5} \int_{0}^{5}\left(\sqrt{5^{2}-x^{2}}\right) d x$
$\\ =\frac{16}{5}\left[\frac{x \sqrt{5^{2}-x^{2}}}{2}+\frac{5^{2}}{2} \sin ^{-1}\left(\frac{x}{5}\right)\right]_{0}^{5} \\ =\frac{16}{5}\left(0-\frac{25 \pi}{4}-0-0\right) \\ =20 \pi \text { sq.units }$
Question 32
The area of the region bounded by the circle $x^{}2 + y^{}2 = 1$ is
A. 2π sq units
B. π sq units
C. 3π sq units
D. 4π sq units
Answer:
B)
The circle $x^{}2 + y^{}2 = 1$
The circle is symmetrical with the x-axis and y-axis
Required Area
$\\ =4 \int_{0}^{1}\left(\sqrt{1-x^{2}}\right) d x \\ {\left[\int \sqrt{a^{2}-x^{2}} d x=\frac{x \sqrt{a^{2}-x^{2}}}{2}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)\right]} \\ =4 \int_{0}^{1}\left(\sqrt{1^{2}-x^{2}}\right) d x \\ =4\left[\frac{x \sqrt{1^{2}-x^{2}}}{2}+\frac{1^{2}}{2} \sin ^{-1}\left(\frac{x}{1}\right)\right]_{0}^{1} \\ =4\left(0-\frac{\pi}{4}-0-0\right) \\ =\pi \text { sq.units }$
Question 33
The area of the region bounded by the curve y = x + 1 and the lines x = 2 and x = 3 is
A. $\frac{7}{2}$ sq units
B. $\frac{9}{2}$ sq units
C. $\frac{11}{2}$ sq units
D. $\frac{13}{2}$ sq units
Answer:
A)
The questions gives y = x + 1 and lines x = 2 and x = 3
Required area
$\\ =\int_{2}^{3}(\mathrm{x}+1) \mathrm{d} \mathrm{x} \\ =\left[\frac{\mathrm{x}^{2}}{2}+\mathrm{x}\right]_{2}^{3} \\ =\left(\frac{9}{2}+3-2-2\right) \\ =\frac{7}{2} \mathrm{sq} . \text { units }$
Question 34
The area of the region bounded by the curve x = 2y + 3 and the y lines. y = 1 and y = –1 is
A. 4 sq units
B. $\frac{3}{2}$ sq units
C. 6 sq units
D. 8 sq units
Answer:
C)
The question mentions the curve x = 2y + 3 and the lines on y axis y = 1 and y = -1
Required area
$\\ =\int_{-1}^{1}(2 y+3) d x \\ =\left[y^{2}+3 y\right]_{-1}^{1} \\ =(1+3-1+3) \\ =6 \text { sq.units } \\$
NCERT Exemplar Class 12 Maths Solutions Chapter 8: Topics
- Introduction
- The area under simple curves
- Area of the region bounded by a line and a curve
- The area between two curves
Our experienced teachers have covered every question and have created the NCERT Exemplar Class 12 Maths chapter 8 solutions for it. The solutions given by us are simple, methodical and step-oriented. We want our students to understand every step so much so that solving questions will come super easily for them.
NCERT Exemplar Class 12 Maths Solutions Chapterwise
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Download EBookImportance of NCERT Exemplar Class 12 Maths Solutions Chapter 8
Class 12 Maths NCERT exemplar solutions Chapter 8 Application of Integrals touches upon an exhaustive explanation of how the integrals can be used to find the area under curves. Several topics are covered, and there are plenty of things that one will learn.
- Some of the major topics that are covered under the NCERT exemplar solutions for Class 12 Maths Chapter 8 are properties of integrals (definite), applications of integrals (indefinite), limits of integration, etc.
- Other than this, several topics encompass the area under curves and parabolas, like the area between two curves and the area between one curve and one line.
- There are also some major rules and formulas covered, like Walli's formula and Leibniz's rule. The NCERT exemplar Class 12 Maths solutions chapter 8 also covers the value of functions like the gamma function and the integral function.
CBSE Class 12th Syllabus: Subjects & Chapters
Select your preferred subject to view the chapters
NCERT Solutions for Class 12 Maths: Chapter Wise
Students can find every NCERT Class 12 Maths Solution in one spot on Careers360. Use the links below to access them.
NCERT Solutions of Class 12 - Subject-wise
Here are the subject-wise links for the NCERT Solutions of Class 12:
- NCERT Solutions for Class 12 Physics
- NCERT Solutions for Class 12 Chemistry
- NCERT Solutions for Class 12 Maths
- NCERT Solutions for Class 12 Biology
NCERT Notes of Class 12 - Subject Wise
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NCERT Books and NCERT Syllabus
To plan studies effectively, students should review the latest syllabus before the academic year starts. Find the updated syllabus links and recommended reference books below.
- NCERT Books Class 12 Maths
- NCERT Syllabus Class 12 Maths
- NCERT Books Class 12
- NCERT Syllabus Class 12
NCERT Exemplar Class 12 Solutions - Subject Wise
Given below are the subject-wise Exemplar Solutions of Class 12 NCERT:
Frequently Asked Questions (FAQs)
Q: What are the important topics in the NCERT Exemplar Class 12 Maths Chapter 8?
A:
The use of Integrals is all about integration to determine the area under curves and two functions. It is concerned with definite integrals to determine exact areas with emphasis on graphical representation to visualize regions that are bounded. Students are instructed to use these concepts in daily applications such as physics, economics, and engineering. The ability to draw curves and determine areas facilitates solving problems effectively. Solving NCERT Exemplar problems makes one effective in problem-solving, hence easy to solve integration-based area problems in examinations and practical applications.
Q: What are the real-life applications of the application of integrals in Class 12?
A:
Physics and Engineering are basically applied sciences to find and compute different significant values like area and volume, the amount of work done within a system, and finding the center of mass in mechanics and electric circuits.
Economics and Business – Helps in finding cost, revenue, and profit functions over time.
Biology and Medicine – Used in population growth models, blood flow analysis, and medical imaging, like CT scans.
Architecture and Design – Helps in designing curved structures, bridges, and roads by calculating surface areas.
Astronomy - Applied in the calculation of planetary motion and the space covered by celestial objects.
Q: Why is Chapter 8 Application of Integrals, important in Class 11 Maths?
A:
Class 12 Maths consists of Chapter 8, Application of Integrals, which is important as it is employed in determining the area under curves and between two curves, which is of practical use in real applications like physics, engineering, economics, and architecture. It is a continuation of the concept of integration and uses definite integrals to solve problems of daily life. It is a very important chapter for competitive examinations such as JEE and NEET, in which questions involving integrals in relation to the area are generally asked. These principles are used to solve intricate problems in curved surfaces, motion, and the study of real-life data and thus is an important topic in higher mathematics.
Q: What are some common mistakes students make in Class 12 Application of Integrals?
A:
Class 12 - Application of Integrals has improper limits of integration, where students get the upper and lower limits wrong, and do not use absolute values, since the area is always positive. Most also confuse curves with limits, which results in an improper setup of integration. Overlooking graphical representation results in misinterpretation of problems, whereas errors in calculation during integration, particularly with trigonometric and logarithmic functions, result in incorrect answers. To prevent these errors, students must practice regularly, thoroughly examine the provided curves, and recheck their calculations to avoid errors.
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If you have 6 subjects with Hindi as an additional subject and you have failed in one compartment subject, your additional subject which is Hindi can be considered pass in the board examination.
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The CBSE Class 10 Computer Applications exam (Set-1) was conducted on 27 February 2026 from 10:30 AM to 12:30 PM as part of the CBSE board exams. The paper included MCQs, very short answer questions, short answers, long answers, and case-study questions based on topics like HTML, networking, internet
Popular CBSE Class 12th Questions
A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
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Option 2)
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| Option 3)
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Option 4)
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An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range
| Option 1)
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Option 2)
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| Option 3)
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Option 4)
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A particle is projected at 600 to the horizontal with a kinetic energy . The kinetic energy at the highest point
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Option 2)
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| Option 3)
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Option 4)
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In the reaction,
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Option 2)
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| Option 3)
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Option 4)
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How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?
| Option 1)
0.02 |
Option 2)
3.125 × 10-2 |
| Option 3)
1.25 × 10-2 |
Option 4)
2.5 × 10-2 |
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