RD Sharma Class 12 Exercise 24.4 Areas of Bounded Regions Solutions Maths

RD Sharma Class 12 Exercise 20.4 Areas under the curve Solutions Maths - Download PDF Free Online

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The Class 12 RD Sharma chapter 20 exercise 20.4 solution deals with the chapter of 'Area of bounded region,' which brings out the concepts of a bounded region which is a very essential concept for students if they have to understand the basic maths in a further road of knowledge. The RD Sharma class 12 exercise 20.4 collects the most critical questions important in this chapter.

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RD Sharma Class 12 Solutions Chapter20 Areas Of Bounded Region - Other Exercise
Areas of Bounded Regions Excercise: 20.4
Area of bounded region exercise 20.4 question 3.2
RD Sharma Chapter-wise Solutions

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

Areas of Bounded Regions Excercise: 20.4

Area of bounded region exercise 20.4 question 1

Answer:

$\frac{32}{3}$ sq.units
Hint:
Find the point of intersection of the parabola and the given line and then find the required region.
Given:
$x=4y-y^{2}$ and $x=2y-3$
Solution:
To find the point at the intersection of the parabola $x=4y-y^{2}$ and the line $x=2y-3$

Let us substitute $x=2y-3$ in the equation of the parabola
$\begin{aligned} &2 y-3=4 y-y^{2} \\ &y^{2}-2 y-3=0 \\ &(y+1)(y-3)=0 \\ &y=-1,3 \end{aligned}$
Therefore, the point of intersection are D(-1,-5) and A(3,3)
The area of the required region ABCDOA,
$A=\int_{-1}^{3}\left | x_{1}-x_{2} \right |dy$ where $x=4y-y^{2}$ and $x=2y-3$
$\begin{aligned} &=\int_{-1}^{3}\left(x_{1}-x_{2}\right) d y \quad\left[\because x_{1}>x_{2}\right] \\ &=\int_{-1}^{3}\left[\left(4 y-y^{2}\right)-(2 y-3)\right] d y \\ \end{aligned}$
$\begin{aligned} &=\int_{-1}^{3}\left[4 y-y^{2}-2 y+3\right] d y \\ &=\int_{-1}^{3}\left[-y^{2}+2 y+3\right] d y \\ \end{aligned}$

$\begin{aligned} &=\left[-3^{2}+3^{2}+9-\frac{1}{3}-1+3\right] \\ &=11-\frac{1}{3} \\ &=\frac{32}{3} s q \cdot \text { units } \end{aligned}$

Area of bounded region exercise 20.4 question 3.1

Answer:

9 sq. units
Hint:
Find the point of intersection of parabola and the given line and then integrate it to find the required region.
Given:
$y=2x-4$ and $y^{2}=4x$
Solution:

Substitute $y=2x-4$ in $y^{2}=4x$
$\begin{aligned} &(2 x-4)^{2}=4 x \\ &4 x^{2}+16-16 x=4 x \\ &4 x^{2}-20 x+16=0 \\ \end{aligned}$
$\begin{aligned} &x^{2}-5 x+4=0 \\ &(x-1)(x-4)=0 \\ &x=1,4 \\ &y=-2,4 \end{aligned}$
Therefore point of intersection are (1,-2) and (4,4)
Using horizontal strips:
The required area is the shaded region
$\begin{aligned} A &=\int_{-2}^{4}\left(x_{1}-x_{2}\right) d y \end{aligned}$ where $\begin{aligned} x_{1}=\frac{y+4}{2} \end{aligned}$ and $\begin{aligned} x_{2}=\frac{y^{2}}{4} \\ \end{aligned}$
$\begin{aligned} &=\int_{-2}^{4}\left[\left(\frac{y+4}{2}\right)-\left(\frac{y^{2}}{4}\right)\right] d y \\ \end{aligned}$

$\begin{aligned} &=\left[\frac{4^{2}}{4}+2(4)-\frac{4^{3}}{12}\right]-\left[\frac{(-2)^{2}}{4}+2(-2)-\frac{(-2)^{3}}{12}\right] \\ \end{aligned}$

Area of bounded region exercise 20.4 question 4

Answer:

18 sq.units
Hint:
Find the point of intersection of parabola and the given line and then integrate it to find the required region.
Given:
x-y=4 and $y^{2}=2x$
Solution:

The parabola $y^{2}=2x$ opens towards the positive x-axis and its focus is $\left ( \frac{1}{2},0 \right )$
The straight line $x-y=4$ passes through (4,0) and (0,-4)
Solving $x-y=4$ and $y^{2}=2x$ , we get
$\begin{aligned} &y^{2}=2(y+4) \\ &y^{2}-2 y-8=0 \\ &(y-4)(y+2)=0 \\ &y=4,-2 \end{aligned}$
So, the points of intersection of given parabola and the line are(8,4) and (2,-2)
Required area,
$\begin{aligned} &=\int_{-2}^{4} x_{\text {line }} d y-\int_{-2}^{4} x_{\text {parabola }} d y \\ \end{aligned}$
$\begin{aligned} &=\int_{-2}^{4}(y+4) d y-\int_{-2}^{4} \frac{y^{2}}{2} d y \\ \end{aligned}$
9 $\begin{aligned} &=\frac{(y+4)^{2}}{2}-\left[\frac{1}{2}\left(\frac{y^{3}}{3}\right)\right]_{-2}^{4} \\ \end{aligned}$
$\begin{aligned} &=\frac{1}{2}(64-6)-\frac{1}{6}[64-(-8)] \\ &=30-12 \\ &=18 \mathrm{sq} \cdot \text { units } \end{aligned}$

Area of bounded region exercise 20.4 question 3.2

Answer:

$\frac{23}{3}$ Hint:
Find the point of intersection of parabola and the given line and then integrate it to find the required region.
Given:
$y=2x-4$ and $y^{2}=4x$
Solution:

Substitute $y=2x-4$ in $y^{2}=4x$
$\begin{aligned} &(2 x-4)^{2}=4 x \\ &4 x^{2}+16-16 x=4 x \\ &4 x^{2}-20 x+16=0 \\ \end{aligned}$
$\begin{aligned} &x^{2}-5 x+4=0 \\ &(x-1)(x-4)=0 \\ &x=1,4 \\ &y=-2,4 \end{aligned}$
Therefore point of intersection are (1,-2) and (4,4)
Using vertical strips:
The required area is the shaded region
$A=\int_{0}^{4}y_{2}dx-\int_{1}^{4}y_{1}dx$ where $y_{2}=2\sqrt{2}$ and $y_{1}=2x-4$
$\begin{aligned} &=\int_{0}^{4}(2 \sqrt{x}) d x-\int_{1}^{4}(2 x-4) d x \\ &=\left[\frac{4}{3} x^{\frac{3}{2}}\right]_{0}^{4}-\left[x^{2}-4 x\right]_{1}^{4} \\ \end{aligned}$
$\begin{aligned} &=\left[\frac{4}{3}(4)^{\frac{3}{2}}\right]-\left[\frac{4}{3}(0)^{\frac{3}{2}}\right]-\left[4^{2}-4(4)\right]-[1-4(1)] \\ \end{aligned}$
$\begin{aligned} &=\left[\frac{32}{3}-0\right]-[0-(1-4)] \\ &=\frac{32}{3}-3 \\ &=\frac{23}{3} s q \cdot \text { units } \end{aligned}$

The RD Sharma class 12th exercise 20.4 consists of a total of 5 questions that are short and easy, covering up the effectively important concepts of this chapter that are profitable for board exams. The concepts covered in the RD Sharma class 12 solution of Area of bounded region exercise 20.4 are-

The line segment of the curve
Area of the special bounded region
Area of the region that is bounded by the curve
Area of the region that is bounded by a parabola

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