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RD Sharma Class 12 Exercise 20.1 Areas of bounded region Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 20.1 Areas of bounded region Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 24, 2022 03:19 PM IST

RD Sharma books are the benchmark of CBSE Maths in India. These books are renowned for their detailed and concept-rich chapters. They cover all aspects of the chapter and are the best medium of preparation for CBSE students. Once students prepare from this book, they can rest assured that they have covered all concepts in detail.

RD Sharma Class 12th Exercise 20.1 of Indefinite Integrals contains 30 questions, 22 of which are Level 1 and eight are Level 2. The Level 1 sums are relatively easier and can be completed swiftly. Level 2 sums, however, require some conceptual understanding and are slightly longer. RD Sharma solutions These questions are based on the area of the region bounded between line and parabola, curve and line, and many more by using integration and also draw a rough sketch of the graph of the function then evaluate the region. Students are suggested to complete this exercise efficiently with the help of the material provided by Career360.

RD Sharma Class 12 Solutions Chapter20 Areas Of Bounded Region - Other Exercise

Areas of Bounded Regions Excercise: 20.1

Areas of Bounded Region exercise 20.1 question 2

Answer: \frac{17}{2}sq\cdot units
Hint: Use - x+ y = 1
Given: Using integration, find area of region bounded by line y-1=x the x-axis and ordinatex=-2 and x=3
Solution:

Herey-1=xis equation of line

We can write

\begin{aligned} &y=x+1 \\\\ &-x+y=1 \end{aligned}

We get\frac{-x}{1}+\frac{y}{1}=1

Consider AB as line intersecting the x-axis at point C(-1,0)

So required area=\text { Area of } C D A C+\text { Area of } C B E C

=\int_{-1}^{3} y d x+\int_{-2}^{-1}-(y) d x

Substituting the value ofy,

\begin{aligned} &=\int_{-1}^{3}(x+1) d x+\int_{-2}^{-1}-(x+1) d x \\\\ &=\left[\frac{x^{2}}{2}+x\right]_{-1}^{3}-\left[\frac{x^{2}}{2}+x\right]_{-2}^{-1} \end{aligned} \left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right]

Substituting the value ofx ,

\begin{aligned} &=\left(\frac{9}{2}+3\right)-\left(\frac{1}{2}-1\right)-\left(\frac{1}{2}-1\right)-(2-2) \\\\ &=\left(\frac{15}{2}+\frac{1}{2}\right)-\left(\frac{-1}{2}\right) \\\\ &=8+\frac{1}{2} \\ &=\frac{17}{2} s q \cdot \text { units } \end{aligned}

Areas of Bounded Region exercise 20.1 question 3

Answer:
\frac{8}{3}a^{2}\; sq\cdot units
Hint:
Find the shaded area
Given:
Find area of region bounded by parabola y^{2}=4 a x and the line x=a
Solution:

We have
x=a … (i)
y^{2}=4 a x … (ii)
Required area = shaded region OBAO
\left[\because y^{2}=4 a x \Rightarrow y=\sqrt{4 a x}\right]
=2(\text { Shaded region } \mathrm{OBCO}) \quad\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right]
\begin{aligned} =& 2 \int_{0}^{a} \sqrt{4 a x} d x \\\\ &=2 \sqrt{4 a} \int_{0}^{a}(x)^{\frac{1}{2}} d x \\\\ &=2 \sqrt{4 a}\left[\frac{(x)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{a} \end{aligned}
\begin{aligned} &=2 \sqrt{4 a}\left[\frac{(x)^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{a} \\\\ &=2 \sqrt{4 a} \times \frac{2}{3}\left[(x)^{\frac{3}{2}}\right]_{0}^{a} \end{aligned}
\begin{aligned} &=\frac{4 \sqrt{4 a}}{3}\left[(a)^{\frac{3}{2}}-(0)^{\frac{3}{2}}\right] \\\\ &=\frac{4 \sqrt{4 a}}{3} \times(\sqrt{a})^{3} \\\\ &=\frac{4 \sqrt{4 a}}{3} \times(\sqrt{a} \times \sqrt{a} \times \sqrt{a}) \end{aligned}
\begin{aligned} &=\frac{4 \sqrt{4 a}}{3} \times a \sqrt{a} \\\\ &=\frac{8 a^{2}}{3} \text { sq.unit } \end{aligned}

Areas of Bounded Region exercise 20.1 question 4

Answer:
\frac{32}{3}sq\cdot units
Hint:
Solve area of integration.
Given:
Find area lying above the x-axis and under parabola y=4x-x^{2}
Solution:

\begin{aligned} &A=\int_{0}^{4}|y| d x\\\\ &A=\int_{0}^{4} y d x \end{aligned} \text { [As } y>0 \text { for } 0,0 \leq x \leq 4|y|=4]
\begin{aligned} &A=\int_{0}^{4}\left(4 x-x^{2}\right) d x \\\\ &A=\left[\frac{4 x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{4} \\\\ &A=32-\frac{64}{3} \\ &A=\frac{32}{3} s q \cdot u n i t \end{aligned}

Areas of Bounded Region exercise 20.1 question 5

Answer:
8\sqrt{3}\; sq\cdot units
Hint:
A=2 \times \text { Area of } O A B O
Given:
Draw rough sketch to indicate the region bounded between curve y^{2}=4 x and line x=3 .Also find the area of region.
Solution:

y^{2}=4 x Represent parabola with vertex at x=3 line parallel toy-axis.
Since y^{2}=4 x is symmetrical axis
Area of corresponding rectangle =\left ( y \right )dx
\begin{aligned} &A=2 \times \text { Area of } O A B O \\\\ &A=2 \int_{0}^{3}|y| d x \\\\ &A=2 \int_{0}^{3} y d x \end{aligned}
\begin{aligned} &A=2 \int_{0}^{3} \sqrt{4 x} d x \\\\ &A=2 \times 2 \int_{0}^{3} \sqrt{x} d x \\\\ &A=4 \int_{0}^{3} \sqrt{x} d x \end{aligned}
\begin{aligned} &A=4\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{3} \\\\ &=\frac{8}{3}\left[(x)^{\frac{3}{2}}\right]_{0}^{3} \\\\ &=\frac{8}{3} \times 3 \sqrt{3} \\\\ &=8 \sqrt{3} s q \cdot \text { unit } \end{aligned} \left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right]

Areas of Bounded Region exercise 20.1 question 6

Answer:
\frac{16}{3}sq\cdot units
Hint:
y=4-x^{2}, x=0, x=2
Given:
Make rough sketch of graph of function y=4-x^{2} ,0\leq x\leq 2 and determine the area enclosed by curve and x-axis ,line x=0 and x=2 .
Solution:

y=4-x^{2} , 0\leq x\leq 2 represent half parabola with vertex \left ( 2,0 \right )
x=2 represent a line parallel to y-axis and cutting axis at \left ( 2,0 \right )
Area required = y dx
\begin{aligned} &A=\text { Area of } O A B O \\\\ &A=\int_{0}^{2}|y| d x \\\\ &A=\int_{0}^{2} y d x \end{aligned}
\begin{aligned} &A=\int_{0}^{2}\left(4-x^{2}\right) d x \\\\ &A=\left[4 x-\frac{x^{3}}{3}\right]_{0}^{2} \end{aligned}
\begin{aligned} &A=8-\frac{8}{3} \\\\ &A=\frac{16}{3} s q \cdot u n i t \quad\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right] \end{aligned}

Areas of Bounded Region exercise 20.1 question 7

Answer:
\frac{2}{3}\left(5^{\frac{3}{2}}-1\right)
Hint:
Where y=\sqrt{x+1} \text { in }[0,4]
Given:
Sketch the graph of y=\sqrt{x+1}in\left [ 0,4 \right ]and determine the area of region enclosed by curve, thex-axisand line x=0,x=4
Solution:

\text { Area of } O A B C O=\int_{0}^{4}|y| d x \quad\quad\quad\quad[y>0 \Rightarrow|y|=y]
\begin{aligned} &A=\int_{0}^{4} y d x \\\\ &A=\int_{0}^{4} \sqrt{x+1} d x \\\\ &A=\int_{0}^{4}(x+1)^{\frac{1}{2}} d x \end{aligned}
\begin{aligned} &A=\left[\frac{(x+1)^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{4} \\\\ &A=\frac{2}{3}\left(5^{\frac{3}{2}}-1\right) s q \cdot \text { unit } \end{aligned} \left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right]

Areas of Bounded Region exercise 20.1 question 8

Answer:
\frac{56}{9}sq\cdot units
Hint:
y=\sqrt{6 x+4} \quad \text { at } x=0, x=2
Given:
Find area under curve y=\sqrt{6 x+4} above x-axis form x=0 to x=2 . Draw sketch of curve also.
Solution:

y=\sqrt{6 x+4} represent a parabola with vertex v=\left(\frac{-2}{3}, 0\right) and symmetric aboutx-axis where x=0 is y-axis
The rectangle move from x=0 to x=2
Consider,
\begin{aligned} &\text { Area } O A B C=\int_{0}^{2}|y| d x \\\\ &A=\int_{0}^{2} \sqrt{6 x+4} d x \\\\ &A=\int_{0}^{2}(6 x+4)^{\frac{1}{2}} d x \end{aligned}
\begin{aligned} &A=\frac{1}{6}\left[\frac{(6 x+4)^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{2} \\\\ &A=\frac{2}{18}\left[(16)^{\frac{3}{2}}-(4)^{\frac{3}{2}}\right] \end{aligned}
\begin{aligned} &A=\frac{2}{18}\left[4^{3}-2^{3}\right] \\\\ &A=\frac{2}{18}[64-8] \\\\ &A=\frac{2}{18}[56] \\\\ &A=\frac{56}{9} s q \cdot \text { unit } \end{aligned}

Areas of Bounded Region exercise 20.1 question 9

Answer:
\frac{4}{3}sq\cdot unit
Hint:
\text { Area of } A B C A=2 \times \text { Area of } A B D A
Given:
Draw the rough sketch of v^{2}+1=x, x \leq 2 .Find area enclosed by curve and line x=2
Solution:

\begin{aligned} &y^{2}+1=x, x \leq 2\\\\ &y^{2}+1=x \Rightarrow y=\sqrt{x-1}\\\\ &\text { Area of } A B C A=2 \times \text { Area of } A B D A\\\\ &A=2 \int_{1}^{2}|y| d x \end{aligned}
\begin{aligned} &A=2 \int_{1}^{2} y d x \quad[y>0 \Rightarrow|y|=y] \\\\ &A=2 \int_{1}^{2} \sqrt{x-1} d x \end{aligned}
\begin{aligned} &A=2\left[\frac{(x-1)^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{2} \\\\ &A=\frac{4}{3}\left[1^{\frac{3}{2}}-0\right] \\\\ &A=\frac{4}{3} s q \cdot \text { unit } \end{aligned}

Areas of Bounded Region exercise 20.1 question 10

Answer:
3\pi \; sq\cdot unit
Hint:
\frac{x^{2}}{4}+\frac{y^{2}}{9}=1
Given:
Draw rough sketch of graph of curve\frac{x^{2}}{4}+\frac{y^{2}}{9}=1and evaluate area of region under curve and abovex-axis
Solution:

Since given equation\frac{x^{2}}{4}+\frac{y^{2}}{9}=1
All power of x and y are even
A = Area of enclosed curve along x-axis
\begin{aligned} &=2 \int_{0}^{2}|y| \mathrm{dx} \\\\ &A=2 \int_{0}^{2} y \mathrm{dx} \\\\ &A=2 \int_{0}^{2} \frac{3}{2} \sqrt{4-x^{2}} d x \end{aligned}
\begin{aligned} &A=3 \int_{0}^{2} \sqrt{4-x^{2}} d x \\\\ &A=3\left[\frac{1}{2} x \sqrt{4-x^{2}}+\frac{1}{2} 4 \sin ^{-1} \frac{x}{2}\right]_{0}^{2} \end{aligned}
\begin{aligned} &A=3\left[0+\frac{1}{2} \times 4 \sin ^{-1} 1\right] \\\\ &A=3 \times \frac{1}{2} \times 4 \times \frac{\pi}{2} \\\\ &A=3 \pi \text { sq.units } \end{aligned}

Areas of Bounded Region exercise 20.1 question 11

Answer:
6\pi \; sq\cdot unit
Hint:
9 x^{2}+4 y^{2}=36
Given:
Sketch the region (x, y): 9 x^{2}+4 y^{2}=36 and find the area of region enclosed by using integration.
Solution:

We have,
9 x^{2}+4 y^{2}=36 .........(i)
\begin{aligned} &4 y^{2}=36-9 x^{2} \\\\ &y^{2}=\frac{9}{4}\left(4-x^{2}\right) \\\\ &y=\frac{3}{2} \sqrt{4-x^{2}} \end{aligned} ...........(ii)
From (i), we get
\frac{x^{2}}{4}+\frac{y^{2}}{9}=1
Since \frac{x^{2}}{4}+\frac{y^{2}}{9}=1 all power of x and y are equal
\begin{aligned} &A=\text { Area of enclosed curve }=4 \int_{0}^{2}|y| d x \\\\ &A=4 \int_{0}^{2} \frac{3}{2} \sqrt{4-x^{2}} d x \\\\ &A=4 \times \frac{3}{2} \int_{0}^{2} \sqrt{4-x^{2}} d x \end{aligned}
\begin{aligned} &A=6 \int_{0}^{2} \sqrt{2^{2}-x^{2}} d x \\\\ &A=6\left[\frac{x}{2} \sqrt{2^{2}-x^{2}}+\frac{1}{2} 2^{2} \sin ^{-1} \frac{x}{2}\right]_{0}^{2} \\\\ &A=6\left[0+\frac{1}{2} 4 \sin ^{-1} 1\right] \end{aligned}
\begin{aligned} &A=6\left[\frac{1}{2} \times 4\left(\frac{\pi}{2}\right)\right] \\\\ &A=6 \pi \mathrm{sq} \cdot \text { unit } \\\\ &{\left[\because \int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c\right] \quad \text { and }\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right]} \end{aligned}

Areas of Bounded Region exercise 20.1 question 12

Answer:
\frac{\pi}{2} \; sq\cdot unit
Hint:
\left[\because \int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c\right]
Given:
Draw a rough sketch of graph of function y=2 \sqrt{1-x^{2}}, x \in[0,1] and evaluate the area enclosed between the curve and x-axis
Solution:

We have
\begin{aligned} &y=2 \sqrt{1-x^{2}} \\\\ &\frac{y}{2}=\sqrt{1-x^{2}} \\\\ &\frac{y^{2}}{4}=1-x^{2} \end{aligned}
\begin{aligned} x^{2}+\frac{y^{2}}{4} &=1 \\\\ \frac{x^{2}}{1}+\frac{y^{2}}{4} &=1 \end{aligned}
Since given equation \frac{x^{2}}{1}+\frac{y^{2}}{4}=1
\begin{aligned} &A=\int_{0}^{1} 2 \sqrt{1-x^{2}} d x \\\\ &A=2 \int_{0}^{1} \sqrt{1-x^{2}} d x \\\\ &A=2\left[\frac{1}{2} x \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x\right]_{0}^{1} \end{aligned}
\begin{aligned} &A=2\left[\frac{1}{2} \sin ^{-1} 1\right] \\\\ &A=\sin ^{-1} 1 \\\\ &A=\frac{\pi}{2} s q \cdot \text { unit } \end{aligned} \left[\because \int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c\right]

Areas of Bounded Region exercise 20.1 question 13

Answer:
\frac{\pi a^{2}}{4} s q \cdot \text { unit }
Hint:
y=\sqrt{a^{2}-x^{2}} \text { at line } x=0, x=a
Given: Determine the area under curve y=\sqrt{a^{2}-x^{2}} include between line x=0 and x=a
Solution:

\begin{aligned} &y=\sqrt{a^{2}-x^{2}} \\\\ &y^{2}=a^{2}-x^{2} \\\\ &x^{2}+y^{2}=a^{2} \end{aligned}
Required area = area of shaded region
\begin{aligned} &=\int_{0}^{a} y d x \\\\ &=\int_{0}^{a} \sqrt{a^{2}-x^{2}} d x \\\\ &=\left[\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right]_{0}^{a} \end{aligned}
\begin{aligned} &=\left[\frac{a}{2} \sqrt{a^{2}-a^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{a}{a}\right]-\left[\frac{0}{2} \sqrt{a^{2}-0}+\frac{a^{2}}{2} \sin ^{-1} \frac{0}{a}\right] \\\\ &=\frac{a^{2}}{2} \sin ^{-1} 1 \\\\ &=\frac{a^{2}}{2} \times \frac{\pi}{2} \\\\ \end{aligned}
=\frac{\pi a^{2}}{4} s q \cdot u n i t

Areas of Bounded Region exercise 20.1 question 15

Answer:
\pi a^{2}\; s q \cdot \text { unit }
Hint:
Use definite integral
Given:
Using definite integral, find area of circle x^{2}+y^{2}=a^{2}
Solution:

x^{2}+y^{2}=a^{2}
Centre=\left ( 0,0 \right )
Radius=a
Hence, OA=OB=Radius=a
A=(a,0),B=(0,a)
Area of circle=4\times area of region OBAO
=4 \int_{0}^{a} y d x
We know that
\begin{aligned} &x^{2}+y^{2}=a^{2} \\\\ &y^{2}=a^{2}-x^{2} \\\\ &y=\pm \sqrt{a^{2}-x^{2}} \end{aligned}
Since AOBA lies in first quadrant, value of y is positive
y=\sqrt{a^{2}-x^{2}}
Now,
Area of circle =4 \int_{0}^{a} \sqrt{a^{2}-x^{2}} d x \quad\left[\because \int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c\right]
\begin{aligned} &=4\left[\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right]_{0}^{a} \\\\ &=4\left[\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{a}{a}\right]-\left[\frac{0}{2} \sqrt{a^{2}-0}+\frac{0^{2}}{2} \sin ^{-1} \frac{0}{a}\right] \end{aligned}
\begin{aligned} &=4\left[0+\frac{a^{2}}{2} \sin ^{-1} 1\right]_{0}^{a} \\\\ &=4 \times \frac{a^{2}}{2} \times \frac{\pi}{2} \\\\ &=\pi a^{2} \text { sq.units } \end{aligned}

Areas of Bounded Region exercise 20.1 question 16

Answer:
\frac{27}{2} s q \cdot \text { units }
Hint:
y=1+|x+1|, x=-2, x=3, y=0
Given:
Using integration, find area of region bounded by following curve making a rough sketch
y=1+|x+1|, x=-2, x=3, y=0
Solution:

We have
y=1+|x+1| Intersect x=-2 at \left ( -2,2 \right ) and x=3 at (3,5)
y=0 is x-axis

y=1+|x+1|

\left\{\begin{array}{cc} 1-(x+1) & x \leq-1 \\ 1+(x+1) & x \geq 1 \\ -x & x \leq-1 \\ x+2 & x \geq 1 \end{array}\right.
Let required area be A since limits on x are given we use horizontal strip to find area
\begin{aligned} &A=\int_{-2}^{3}|y| d x \\\\ &A=\int_{-2}^{-1}|y| d x+\int_{-1}^{3}|y| d x \\\\ &A=\int_{-2}^{-1}-x d x+\int_{-1}^{3}(x+2) d x \end{aligned}
\begin{aligned} &A=-\left[\frac{x^{2}}{2}\right]_{-2}^{-1}+\left[\frac{x^{2}}{2}+2 x\right]_{-1}^{3} \\\\ &A=-\left[\frac{1}{2}-\frac{4}{2}\right]+\left[\frac{9}{2}+6-\frac{1}{2}+2\right] \\\\ &A=\frac{3}{2}+8+\frac{8}{2} \end{aligned}
\begin{aligned} &A=\frac{3}{2}+8+4 \\\\ &A=\frac{27}{2} s q \cdot u n i t \end{aligned} \quad\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right]

Areas of Bounded Region exercise 20.1 question 17

Answer:
\frac{9}{2}\; s q \cdot unit
Hint:
Break the limit and find the value of integral.
\int_{0}^{1}-(x-5) d x
Given:
Sketch the graph y=\left | x-5 \right |
Evaluate\int_{0}^{1}|x-5| d x. What the value of integral represents on graph
Solution:

\begin{aligned} &\int_{0}^{1}|x-5| d x=\int_{0}^{1}-(x-5) d x \\\\ &=\left[\frac{-x^{2}}{2}+5 x\right]_{0}^{1} \\\\ &=\left[\frac{-1}{2}+5\right] \end{aligned}
=\frac{9}{2}\; s q \cdot unit \left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right]

Areas of Bounded Region exercise 20.1 question 18

Answer:
9\; s q \cdot \text { units }
Hint:
Break the limit and find integral
Given:
y=\left | x+3 \right | and evaluate \int_{-6}^{0}|x+3| d x What represent of graph
Solution:

Let's draw graph,y=\left | x+3 \right |
y=|x+3|=\left\{\begin{array}{cc} x+3 & \text { for } x+3 \geq 0 \\ -(x+3) & \text { for } x+3<0 \end{array}\right.
\left\{\begin{array}{c} x+3 \quad \text { for } x \geq-3 \\ -(x+3) \text { for } x+3<-3 \end{array}\right.
\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right]
Required area
\begin{aligned} &=\int_{-6}^{0}|x+3| d x \\\\ &=\int_{-6}^{-3}|x+3| d x-=\int_{-3}^{0}|x+3| d x \\\\ &=\int_{-6}^{-3}-(x+3) d x-\int_{-3}^{0}(x+3) d x \end{aligned}
\begin{aligned} &=\left[\frac{-x^{2}}{2}-3 x\right]_{-6}^{-3}+\left[\frac{x^{2}}{2}+3 x\right]_{-3}^{0} \\\\ &=\left[\frac{-(-3)^{2}}{2}-3(-3)\right]_{-6}^{-3}-\left[\frac{-(-6)^{2}}{2}-3(-6)\right]+\left[\frac{0^{2}}{2}+3(0)\right]-\left[\frac{(-3)^{2}}{2}+3(-3)\right] \end{aligned}
\begin{aligned} &=\frac{-9}{2}-(-9)-\frac{-36}{2}-(-18)+\left[0+\left(\frac{-9}{2}+9\right)\right] \\\\ &=\frac{-9}{2}+9+0-\frac{9}{2}+9 \\\\ &=-9+18 \\\\ &=9 \text { sq.units } \end{aligned}

Areas of Bounded Region exercise 20.1 question 19

Answer:
9\; s q \cdot { units }
Hint:
Find integral \int_{-4}^{2}|x+1| d x
Given:
y=|x+1| Evaluate \int_{-4}^{2}|x+1| what does the value of integral represented on graph
Solution:

We have,
y=|x+1| intersect x = -4 and x = 2 at ( -4,3 ) and ( 2,3 ).
Now,
y=|x+1|
=\left\{\begin{array}{c} (x+1) \text { for all } x>-1 \\ -(x+1) \text { for all } x<-1 \end{array}\right.
Integral represents the area enclosed between x = -4 and x = 2.
\begin{aligned} A &=\int_{-4}^{2}|y| d x \\\\ A &=\int_{-4}^{-1}|y| d x+\int_{-1}^{2}|y| d x \\\\ A &=\int_{-4}^{-1}-(x+1) d x+\int_{-1}^{2}(x+1) \end{aligned}
\begin{aligned} &A=-\left[\frac{x^{2}}{2}+x\right]_{-4}^{-1}+\left[\frac{x^{2}}{2}+x\right]_{-1}^{2} \\\\ &A=-\left[\frac{1}{2}-1-\frac{16}{2}+4\right]+\left[\frac{4}{2}+2-\frac{1}{2}+1\right] \end{aligned}
\begin{aligned} &A=-\left[3-\frac{15}{2}\right]+\left[5-\frac{1}{2}\right] \\\\ &A=-3+\frac{15}{2}+5-\frac{1}{2} \\\\ &A=\frac{-6+15+10-1}{2} \end{aligned}
A=9\; s q \cdot { units }

Areas of Bounded Region exercise 20.1 question 20

Answer:
3+16\log 2\; s q \cdot { unit }
Hint:
Use the concept of definite integrals.
Given:
Find area of region bounded by curvex y-3 x-2 y-10=0, x-a x i sand line x=3, x=4
Solution:
Given curve,
\begin{aligned} &x y-3 x-2 y-10=0 \\\\ &x y-2 y=3 x+10 \\\\ &y(x-2)=3 x+10 \\\\ &y=\frac{3 x+10}{x-2} \end{aligned}

Area of bounded curve x y-3 x-2 y-10=0, x-a x i s and line x=3, x=4
\begin{aligned} &\int_{3}^{4}|y| d x=\int_{3}^{4}\left(\frac{3 x+10}{x-2}\right) d x \\\\ &=\int_{3}^{4}\left(\frac{3 x-6+16}{x-2}\right) d x \\\\ &=\int_{3}^{4}\left(3+\frac{16}{x-2}\right) d x \end{aligned}
\begin{aligned} &=[3 x+16 \log |x-2|]_{3}^{4} \\\\ &=12+16 \log |2|-9-16 \log |1| \\\\ &=16 \log 2+3 \text { sq.unit } \end{aligned}

Areas of Bounded Region exercise 20.1 question 21

Answer:
\frac{\pi }{2}(\pi +2) s q \cdot { unit }
Hint:
use definite integral.
Given:
Draw rough sketch of curve y=\frac{\pi}{2}+2 \sin ^{2} x Find area between x-axis the curve and ordinate x=0,x=\pi
Solution:

x
0
\frac{\pi }{6}
\frac{\pi }{2}
\frac{5\pi }{6}
\pi
sin \; x
0
\frac{1}{2}
1
\frac{1}{2}

0

2\sin ^{2}x
1.57
2.07
3.57
2.07
1.5
y=\frac{\pi}{2}+2 \sin ^{2} x is an arc cutting y-axis at \left ( 1.57,0 \right )

x=\piat(\pi ,1.57)

Area of shaded region
\begin{aligned} &A=\int_{0}^{\pi}|y| d x \\\\ &A=\int_{0}^{\pi} y d x \\\\ &A=\int_{0}^{\pi}\left[\frac{\pi}{2}+2 \sin ^{2} x\right] d x \end{aligned}
\begin{aligned} &A=\int_{0}^{\pi}\left[\frac{\pi}{2}+2\left(\frac{1-\cos 2 x}{2}\right)\right] d x \\\\ &A=\int_{0}^{\pi}\left[\frac{\pi}{2}\right] d x+\int_{0}^{\pi}[(1-\cos 2 x)] d x \\\\ &A=\frac{\pi}{2}[x]_{0}^{\pi}+\left[x-\frac{\sin 2 x}{2}\right]_{0}^{\pi} \end{aligned}
\begin{aligned} &A=\frac{\pi}{2}[\pi]+\left[\pi-\frac{\sin 2 \pi}{2}-0\right] \\\\ &A=\pi\left(\frac{\pi}{2}+1\right) \\\\ &A=\pi\left(\frac{\pi+2}{2}\right) \end{aligned}
A=\frac{\pi }{2}(\pi +2) s q \cdot{ unit }

Areas of Bounded Region exercise 20.1 question 22

Answer:
\frac{3\pi }{2} s q \text { unit }
Hint:
You know about integral of \sin ^{2}x
Given:
Draw rough curve y=\frac{x}{\pi}+2 \sin ^{2} x . Find area between x-axis with ordinate x=0, x=\pi
Solution:

x
0
\frac{\pi }{6}
\frac{\pi }{2}
\frac{5\pi }{6}
\pi
\sin x
0
\frac{1}{2}
1
\frac{1}{2}
0
\frac{x}{\pi }+2\sin ^{2}x
0
\frac{2}{3}
\frac{5}{2}
\frac{4}{3}
1
A=\int_{0}^{\pi}|y| d x \quad\quad\quad\quad[y>0 \Rightarrow|y|=y]
\begin{aligned} &A=\int_{0}^{\pi}\left(\frac{x}{\pi}+2 \sin ^{2} x\right) d x \\\\ &A=\frac{1}{\pi} \int_{0}^{\pi}(x) d x+2 \int_{0}^{\pi}\left(\sin ^{2} x\right) d x \\\\ &A=\frac{1}{\pi}\left[\frac{x^{2}}{2}\right]_{0}^{\pi}+2\left[\frac{x}{2}-\frac{1}{2} \sin x \cos x\right] \end{aligned}
\begin{aligned} &A=\frac{\pi^{2}}{2 \pi}+\frac{2}{2}\left[\pi-\frac{1}{2} \sin x \cos x-0\right] \\\\ &A=\frac{\pi}{2}+\pi \\\\ &A=\frac{3 \pi}{2} s q \cdot u n i t \end{aligned}

Areas of Bounded Region exercise 20.1 question 23

Answer:
4\; s q \cdot { unit }
Hint:
Curve, y=\cos x
Given:
Find area by curve y=\cos x and ordinate x=0 and x=2\pi
Solution:

\begin{aligned} &A=\int_{0}^{2 \pi}|y| d x \\\\ &A=\int_{0}^{\frac{\pi}{2}}|y| d x+\int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}}|y| d x+\int_{\frac{3 \pi}{2}}^{2 \pi}|y| d x \end{aligned}
\begin{aligned} &A=\int_{0}^{\frac{\pi}{2}} y d x+\int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}}-y d x+\int_{\frac{3 \pi}{2}}^{2 \pi} y d x \\\\ &A=\int_{0}^{\frac{\pi}{2}} \cos x d x+\int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}}-\cos x d x+\int_{\frac{3 \pi}{2}}^{2 \pi} \cos x d x \end{aligned}
\begin{aligned} &A=[\sin x]_{0}^{\frac{\pi}{2}}+[-\sin x]_{\frac{\pi}{2}}^{\frac{3 \pi}{2}}+[\sin x]_{ \frac{3 \pi}{2}}^{2\pi} \\\\ &A=1+[1+1]+0-(-1) \\\\ &A=4 \; \mathrm{sq} \cdot \mathrm{unit} \end{aligned}

Areas of Bounded Region exercise 20.1 question 24

Answer:
y=\sin x, y=\sin 2 x \text { is in ratio } 2: 3
Hint:
Use two graph of \sin x.
Given:
Show that area under curve y=\sin x, y=\sin 2 x between x=0 and x=\frac{\pi }{3} are in ratio 2:3
Solution:


Area of graph 1 :
A_{1}=\int_{0}^{\frac{\pi}{3}}|y| d x \quad\quad\quad\quad[y>0 \Rightarrow|y|=y]
\begin{aligned} &A_{1}=\int_{0}^{\frac{\pi}{3}} y d x \\\\ &A_{1}=\int_{0}^{\frac{\pi}{3}} \sin x d x \\\\ &A_{1}=[-\cos x]_{0}^{\frac{\pi}{3}} \end{aligned}
\begin{aligned} &A_{1}=\left[-\cos \frac{\pi}{3}+\cos 0\right] \\\\ &A_{1}=-\frac{1}{2}+1 \\\\ &A_{1}=\frac{1}{2} \end{aligned}
Area of graph 2 : ............(i)
A_{2}=\int_{0}^{\frac{\pi}{3}}|y| d x \quad\quad\quad\quad[y>0 \Rightarrow|y|=y]
\begin{aligned} &A_{2}=\int_{0}^{\frac{\pi}{3}} y d x \\\\ &A_{2}=\int_{0}^{\frac{\pi}{3}} \sin 2 x d x \\\\ &A_{2}=\int_{0}^{\frac{\pi}{3}} 2 \sin x \cos x d x \end{aligned}
\begin{aligned} &A_{2}=2\left[\frac{-1}{4} \cos 2 x\right]_{0}^{\frac{\pi}{3}} \\\\ &A_{2}=\frac{1}{2}\left[-\cos 2 \frac{\pi}{3}+\cos 0\right] \end{aligned}
\begin{aligned} &A_{2}=\frac{1}{2}\left[1+\frac{1}{2}\right] \\\\ &A_{2}=\frac{1}{2}\left[\frac{3}{2}\right] \\\\ &A_{2}=\frac{3}{4} \end{aligned} ............(ii)
From (i) and (ii)
\frac{A_{1}}{A_{2}}=\frac{\frac{1}{2}}{\frac{3}{4}}=\frac{2}{3}
Thus area of curve y=\sin x, y=\sin 2 x for x=0 and x=\frac{\pi }{3} are in ratio 2:3

Areas of Bounded Region exercise 20.1 question 25

Answer:
Each equal to\frac{\pi }{2} s q \text { unit }
Hint:
Use \sin ^{2} x, \cos ^{2} x
Given:
Compare area under curvey=\cos ^{2} x, y=\sin ^{2} x betweenx=0, x=\pi
Solution:
x
0
\frac{\pi }{6}
\frac{\pi }{4}
\frac{\pi }{3}
\frac{\pi }{2}
\cos ^{2}x
1
0.5
0.75
0.25
0
\sin ^{2}x
0
0.25
0.5
0.75
1

Apply reduction formula,
\begin{aligned} &\int \cos ^{n} x d x=\frac{n-1}{n} \int \cos ^{n-2} x d x+\frac{\cos ^{n-1} x \sin x}{n} \\\\ &A_{1}=2\left[\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} d x+\frac{\sin x \cos x}{2}\right]_{\frac{\pi}{2}}^{\pi} \end{aligned}
\begin{aligned} &A_{1}=[x+\sin x \cos x]_{\frac{\pi}{2}}^{\pi} \\\\ &A_{1}=[\pi+\sin \pi \cos \pi]-\left[\frac{\pi}{2}+\sin \frac{\pi}{2} \cos \frac{\pi}{2}\right] \\\\ &A_{1}=\frac{\pi}{2} \end{aligned}
Apply reduction formula,
\int \sin ^{n} x d x=\frac{n-1}{n} \int \sin ^{n-2} x d x+\frac{\sin ^{n-1} x \cos x}{n}
\begin{aligned} &A_{2}=\left[\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} d x-\frac{\sin x \cos x}{2}\right]_{0}^{\pi} \\\\ &A_{2}=\frac{1}{2}[x-\sin x \cos x]_{0}^{\pi} \end{aligned}
\begin{aligned} &A_{2}=\frac{1}{2}[\pi-\sin \pi \cos \pi]-\frac{1}{2}[0+\sin 0 \cos 0] \\\\ &A_{2}=\frac{\pi}{2} \\\\ &A_{1}=A_{2} \end{aligned}

Areas of Bounded Region exercise 20.1 question 26

Answer:
a b\left\{e \sqrt{1-e^{2}}+\sin ^{-1} e\right\}
Hint:
Use ellipse formula
Given:
Find area bounded by ellipse \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 and ordinate x=ae and x=0 whereb^{2}=a^{2}\left(1-e^{2}\right) and e< 1
Solution:

Required area= Area of region
= Area ofBORQSP
\begin{aligned} &=2 \times \text { Area OBPS } \\\\ &=2 \int_{0}^{a e} y d x \end{aligned}
We know that,
\begin{aligned} &\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \\\\ &\frac{y^{2}}{b^{2}}=\frac{a^{2}-x^{2}}{a^{2}} \\\\ &y^{2}=\frac{b^{2}}{a^{2}}\left(a^{2}-x^{2}\right) \end{aligned}\begin{aligned} &y=\pm \sqrt{\frac{b^{2}}{a^{2}}\left(a^{2}-x^{2}\right)} \\\\ &y=\pm \frac{b}{a} \sqrt{a^{2}-x^{2}} \end{aligned}

Since OBPSin first quadrant, value ofy is positive

\begin{aligned} &y=\frac{b}{a} \sqrt{a^{2}-x^{2}} \\\\ &=2 \int_{0}^{a e} y d x \end{aligned}
Required area =
\begin{aligned} &=2 \int_{0}^{a e} \frac{b}{a} \sqrt{a^{2}-x^{2}} d x \\\\ \end{aligned}
=\frac{2 b}{a} \int_{0}^{a e} \sqrt{a^{2}-x^{2}} d x \quad\left[\because \int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c\right]
\begin{aligned} &=\frac{2 b}{a}\left[\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right]_{0}^{a e} \\\\ &=\frac{2 b}{a}\left[\frac{a e}{2} \sqrt{a^{2}-(a e)^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{a e}{a}\right]-\frac{2 b}{a}\left[\frac{0}{2} \sqrt{a^{2}-0}+\frac{a^{2}}{2} \sin ^{-1} \frac{0}{a}\right] \end{aligned}
\begin{aligned} &=\frac{2 b}{a}\left[\frac{a e}{2} a \sqrt{1-e^{2}}+\frac{a^{2}}{2} \sin ^{-1} e-0\right] \\\\ &=\frac{2 b}{a}\left[\frac{a^{2}}{2} e \sqrt{1-e^{2}}+\frac{a^{2}}{2} \sin ^{-1} e\right] \end{aligned}
\begin{aligned} &=\frac{2 b}{a} \times \frac{a^{2}}{2}\left[e \sqrt{1-e^{2}}+\sin ^{-1} e\right] \\\\ &=a b\left[e \sqrt{1-e^{2}}+\sin ^{-1} e\right] \end{aligned}

Areas of Bounded Region exercise 20.1 question 27

Answer:
\frac{a^{2}}{12}(4 \pi-3 \sqrt{3}) { sq } \cdot { unit }
96\; s q \cdot { unit }
Hint:
Use definite integrals.
Given:
Find area of circle x^{2}+y^{2}=a^{2} cut by the line x=\frac{a}{2}
Solution:

By solving equation \frac{a}{2}, \frac{\sqrt{3} a}{2}and \frac{a}{2}, \frac{-\sqrt{3} a}{2}
Hence form diagram, we get
Required area =2 \times \text { Area of } \mathrm{AOB}
\begin{aligned} &=2 \int_{a / 2}^{a} \sqrt{a^{2}-x^{2}} d x \\\\ &=2\left[\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right]_{a / 2}^{a} \end{aligned}
\begin{aligned} &=2\left[\frac{a^{2}}{2}\left(\frac{\pi}{2}\right)-\frac{a}{4}\left(\frac{a \sqrt{3}}{2}\right)-\frac{a^{2}}{2}\left(\frac{\pi}{6}\right)\right] \\\\ &=\frac{a^{2}}{12}[6 \pi-3 \sqrt{3}-2 \pi] \\\\ &=\frac{a^{2}}{12}[4 \pi-3 \sqrt{3}] { sq\;. unit } \end{aligned}

Areas of Bounded Region exercise 20.1 question 28

Answer:
\frac{56}{3}a^{2}\; s q\cdot { unit }
Hint:
Use definite integral.
Given:
Find area of curve x=a t^{2}, y=2 a t between ordinate t=1 and t=2
Solution:

Given x=a t^{2}, y=2 a t
Required area=2 \times \text { Area of } A B C D
\begin{aligned} &=2 \int_{a}^{4 a} y d x \\\\ &=2 \times 2 \int_{a}^{4 a} \sqrt{a x} d x \end{aligned}
\begin{aligned} &=8 \sqrt{a}\left[\frac{(x)^{\frac{3}{2}}}{\frac{3}{2}}\right]_{a}^{4 a} \\\\ &=\frac{56}{3} a^{2}\; s q \cdot u n i t \end{aligned}

Areas of Bounded Region exercise 20.1 question 29

Answer:
6\pi \; s q \cdot { unit }
Hint:
Use x=3 \cos t, y=2 \sin t
Given:
Find area enclosed by curve x=3 \cos t, y=2 \sin t
Solution:

The given curve x = 3cot\; t, y = 2sin\; t represents the parametric equation of ellipse.
Eliminating the parameter t, we get,
\frac{x^{2}}{9}+\frac{y^{2}}{4}=\cos ^{2} t+\sin ^{2} t=1
This represent the Cartesian equation of the ellipse with centre \left ( 0,0 \right ). The co ordinates of the vertices are \left ( \pm 3,0 \right ) and \left ( 0,\pm 2 \right ).
\therefore Required area = Area of the shaded region
= Area of regionOABO
\begin{aligned} &=4 \times \int_{0}^{3} y_{\text {ellipse }} d x \\\\ &=4 \times \int_{0}^{3} \sqrt{4\left(1-\frac{x^{2}}{9}\right)} d x \\\\ &=4 \times \frac{2}{3} \int_{0}^{3} \sqrt{9-x^{2}} d x \end{aligned}
\begin{aligned} &=\frac{8}{3}\left(\frac{x}{2} \sqrt{9-x}+\frac{9}{2} \sin ^{-1} \frac{x}{3}\right)_{0}^{3} \\\\ &=\frac{8}{3}\left[\left(0+\frac{9}{2} \sin ^{-1} 1\right)-(0+0)\right] \\\\ &=\frac{8}{3} \times \frac{9}{2} \times \frac{\pi}{2} \end{aligned}
=6\pi \; s q \cdot { unit }

Areas of Bounded Region exercise 20.1 question 30

Answer:
16^{\frac{1}{3}}
Hint:
Use integration.
Given:
If area between curve x=y^{2} and x=4 divide two equal part of line x=a .find using integration.
Solution:

Given curve

y^{2}=x
Let AB represent line x=a
CD represent line x=4
Since line x=a divide the region in two equal parts
Area of OBA= Area ofABCD
\begin{aligned} &2 \int_{0}^{a} y d x=2 \int_{a}^{4} y d x \\\\ &\int_{0}^{a} y d x=\int_{a}^{4} y d x \end{aligned} .............(i)
Now, y^{2}=x \Rightarrow y=\pm \sqrt{x}
y=\sqrt{x}
From (i)
\int_{0}^{a} y d x=\int_{a}^{4} y d x
\begin{aligned} &\int_{0}^{a} \sqrt{x} d x=\int_{a}^{4} \sqrt{x} d x \\\\ &\left(\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right)_{0}^{a}=\left(\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right)_{a}^{4} \end{aligned}
\begin{aligned} &{\left[x^{\frac{3}{2}}\right]_{0}^{a}=\left[x^{\frac{3}{2}}\right]_{a}^{4}} \\\\ &a^{\frac{3}{2}}=4^{\frac{3}{2}}-a^{\frac{3}{2}} \\\\ &2 a^{\frac{3}{2}}=4^{\frac{3}{2}} \end{aligned}
Take \frac{2}{3}^{t h}root on both sides
\begin{aligned} &2^{\frac{2}{3}} \cdot a^{\frac{3}{2} \cdot \frac{2}{3}}=4^{\frac{3}{2} \frac{2}{3}} \\\\ &2^{\frac{2}{3}} \cdot a=4 \\\\ &a=\frac{2^{2}}{2^{\frac{2}{3}}} \end{aligned}
\begin{aligned} &a=2^{2-\frac{2}{3}}=2^{\frac{6-2}{3}} \\\\ &a=2^{\frac{4}{3}} \\\\ &a=\left(2^{2}\right)^{\frac{2}{3}}=4^{\frac{2}{3}} \end{aligned}
\begin{aligned} &=\left(4^{2}\right)^{\frac{1}{3}}=16^{\frac{1}{3}} \\\\ &a=16^{\frac{1}{3}} \end{aligned}


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RD Sharma books are more detailed and contain exam-oriented questions. NCERT books are useful for basic education, but they don't get to the level of RD Sharma in the case of content. 

2. Can I solve NCERT questions after preparing this material?

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3. What is a bounded region?

Any curve or figure on a graph that is bounded on all sides is called a bounded region. It can also be an outcome of the intersection of two curves. For more information, check Class 12 RD Sharma Chapter 20 Exercise 20.1 Solutions.

4. How to calculate the area of a bounded region?

The area of a bounded region can be calculated using the integral of its function after applying the horizontal or vertical limits of the region. To know more, refer to RD Sharma Class 12 Solutions Areas of Bounded Region Ex 20.1.

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Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems. 

2 Jobs Available
Naval Architect

A Naval Architect is a professional who designs, produces and repairs safe and sea-worthy surfaces or underwater structures. A Naval Architect stays involved in creating and designing ships, ferries, submarines and yachts with implementation of various principles such as gravity, ideal hull form, buoyancy and stability. 

2 Jobs Available
Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
Veterinary Doctor
5 Jobs Available
Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
Speech Therapist
4 Jobs Available
Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth. 

4 Jobs Available
Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
Audiologist

The audiologist career involves audiology professionals who are responsible to treat hearing loss and proactively preventing the relevant damage. Individuals who opt for a career as an audiologist use various testing strategies with the aim to determine if someone has a normal sensitivity to sounds or not. After the identification of hearing loss, a hearing doctor is required to determine which sections of the hearing are affected, to what extent they are affected, and where the wound causing the hearing loss is found. As soon as the hearing loss is identified, the patients are provided with recommendations for interventions and rehabilitation such as hearing aids, cochlear implants, and appropriate medical referrals. While audiology is a branch of science that studies and researches hearing, balance, and related disorders.

3 Jobs Available
Hospital Administrator

The hospital Administrator is in charge of organising and supervising the daily operations of medical services and facilities. This organising includes managing of organisation’s staff and its members in service, budgets, service reports, departmental reporting and taking reminders of patient care and services.

2 Jobs Available
Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

4 Jobs Available
Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
Videographer
2 Jobs Available
Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

2 Jobs Available
Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

5 Jobs Available
Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

3 Jobs Available
Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
Linguist

Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning). 

Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian

2 Jobs Available
Public Relation Executive
2 Jobs Available
Travel Journalist

The career of a travel journalist is full of passion, excitement and responsibility. Journalism as a career could be challenging at times, but if you're someone who has been genuinely enthusiastic about all this, then it is the best decision for you. Travel journalism jobs are all about insightful, artfully written, informative narratives designed to cover the travel industry. Travel Journalist is someone who explores, gathers and presents information as a news article.

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
QA Manager
4 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Production Manager
3 Jobs Available
Merchandiser
2 Jobs Available
QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

2 Jobs Available
Metallurgical Engineer

A metallurgical engineer is a professional who studies and produces materials that bring power to our world. He or she extracts metals from ores and rocks and transforms them into alloys, high-purity metals and other materials used in developing infrastructure, transportation and healthcare equipment. 

2 Jobs Available
Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

4 Jobs Available
QA Manager
4 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
ITSM Manager
3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
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