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

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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

• Chapter 20 - Areas Of Bounded Region - Ex-20.2

• Chapter 20- Areas Of Bounded Region - Ex-20.3

• Chapter 20- Areas Of Bounded Region - Ex-20.4

• Chapter 20 - Areas Of Bounded Region - Ex-MCQ

• Chapter 20 - Areas Of Bounded Region - Ex-FBQ

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 ordinate$x=-2$ and $x=3$
Solution:

Here$y-1=x$is equation of line

We can write

_{$\begin{array}{rl}& y=x+1\\ \\ & -x+y=1\end{array}$}

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_{$=\text{ Area of }CDAC+\text{ Area of }CBEC$}

_{$={\int }_{-1}^{3}ydx+{\int }_{-2}^{-1}-(y)dx$}

Substituting the value of_$y$,

_{$\begin{array}{rl}& ={\int }_{-1}^3(x+1)dx+{\int }_{-2}^{-1}-(x+1)dx\\ \\ & ={[\frac{x^2}{2}+x]}_{-1}^3-{[\frac{x^2}{2}+x]}_{-2}^{-1}\end{array}$ $[\because {\int }_a^b{x}^{n}dx={\left[\frac{x^{n+1}}{n+1}\right]}_a^b]$}

Substituting the value of_$x$ ,

$\begin{array}{rl}& =(\frac{9}{2}+3)-(\frac{1}{2}-1)-(\frac{1}{2}-1)-(2-2)\\ \\ & =(\frac{15}{2}+\frac{1}{2})-\left(\frac{-1}{2}\right)\\ \\ & =8+\frac{1}{2}\\ & =\frac{17}{2}sq\cdot \text{ units }\end{array}$

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 $y^2=4ax$ and the line $x=a$
Solution:

We have
$x=a$ … (i)
$y^2=4ax$ … (ii)
Required area $=$ shaded region $OBAO$
$[\because y^2=4ax\Rightarrow y=\sqrt{4ax}]$
$=2(\text{ Shaded region }\mathrm{OBCO})[\because {\int }_a^b{x}^{n}dx={\left[\frac{x^{n+1}}{n+1}\right]}_a^b]$
$\begin{array}{rl}=& 2{\int }_0^a\sqrt{4ax}dx\\ \\ & =2\sqrt{4a}{\int }_0^a(x{)}^{\frac{1}{2}}dx\\ \\ & =2\sqrt{4a}{\left[\frac{(x{)}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_0^a\end{array}$
$\begin{array}{rl}& =2\sqrt{4a}{\left[\frac{(x{)}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^a\\ \\ & =2\sqrt{4a}\times \frac{2}{3}{[(x{)}^{\frac{3}{2}}]}_0^a\end{array}$
$\begin{array}{rl}& =\frac{4\sqrt{4a}}{3}[(a{)}^{\frac{3}{2}}-(0{)}^{\frac{3}{2}}]\\ \\ & =\frac{4\sqrt{4a}}{3}\times (\sqrt{a}{)}^3\\ \\ & =\frac{4\sqrt{4a}}{3}\times (\sqrt{a}\times \sqrt{a}\times \sqrt{a})\end{array}$
$\begin{array}{rl}& =\frac{4\sqrt{4a}}{3}\times a\sqrt{a}\\ \\ & =\frac{8a^2}{3}\text{ sq.unit }\end{array}$

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:

$\begin{array}{rl}& A={\int }_0^4|y|dx\\ \\ & A={\int }_0^{4}ydx\end{array}$ $\text{ [As }y>0\text{ for }0,0\le x\le 4|y|=4]$
$\begin{array}{rl}& A={\int }_0^4(4x-x^2)dx\\ \\ & A={[\frac{4x^2}{2}-\frac{x^3}{3}]}_0^4\\ \\ & A=32-\frac{64}{3}\\ & A=\frac{32}{3}sq\cdot unit\end{array}$

Areas of Bounded Region exercise 20.1 question 5

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

$y^2=4x$ Represent parabola with vertex at $x=3$ line parallel to$y-axis$.
Since $y^2=4x$ is symmetrical axis
Area of corresponding rectangle $=\left(y\right)dx$
$\begin{array}{rl}& A=2\times \text{ Area of }OABO\\ \\ & A=2{\int }_0^3|y|dx\\ \\ & A=2{\int }_0^{3}ydx\end{array}$
$\begin{array}{rl}& A=2{\int }_0^3\sqrt{4x}dx\\ \\ & A=2\times 2{\int }_0^3\sqrt{x}dx\\ \\ & A=4{\int }_0^3\sqrt{x}dx\end{array}$
$\begin{array}{rl}& A=4{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^3\\ \\ & =\frac{8}{3}{[(x{)}^{\frac{3}{2}}]}_0^3\\ \\ & =\frac{8}{3}\times 3\sqrt{3}\\ \\ & =8\sqrt{3}sq\cdot \text{ unit }\end{array}$ $[\because {\int }_a^b{x}^{n}dx={\left[\frac{x^{n+1}}{n+1}\right]}_a^b]$

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 $y=4-x^2$ ,$0\le x\le 2$ and determine the area enclosed by curve and $x-axis$ ,line $x=0$ and $x=2$ .
Solution:

$y=4-x^2$ , $0\le x\le 2$ represent half parabola with vertex $(2,0)$
$x=2$ represent a line parallel to $y-axis$ and cutting axis at $(2,0)$
Area required $=ydx$
$\begin{array}{rl}& A=\text{ Area of }OABO\\ \\ & A={\int }_0^2|y|dx\\ \\ & A={\int }_0^{2}ydx\end{array}$
$\begin{array}{rl}& A={\int }_0^2(4-x^2)dx\\ \\ & A={[4x-\frac{x^3}{3}]}_0^2\end{array}$
$\begin{array}{rl}& A=8-\frac{8}{3}\\ \\ & A=\frac{16}{3}sq\cdot unit[\because {\int }_a^b{x}^{n}dx={\left[\frac{x^{n+1}}{n+1}\right]}_a^b]\end{array}$

Areas of Bounded Region exercise 20.1 question 7

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

$\text{ Area of }OABCO={\int }_0^4|y|dx[y>0\Rightarrow |y|=y]$
$\begin{array}{rl}& A={\int }_0^{4}ydx\\ \\ & A={\int }_0^4\sqrt{x+1}dx\\ \\ & A={\int }_0^4(x+1{)}^{\frac{1}{2}}dx\end{array}$
$\begin{array}{rl}& A={\left[\frac{(x+1{)}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^4\\ \\ & A=\frac{2}{3}(5^{\frac{3}{2}}-1)sq\cdot \text{ unit }\end{array}$ $[\because {\int }_a^b{x}^{n}dx={\left[\frac{x^{n+1}}{n+1}\right]}_a^b]$

Areas of Bounded Region exercise 20.1 question 8

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

$y=\sqrt{6x+4}$ represent a parabola with vertex $v=(\frac{-2}{3},0)$ and symmetric about$x-axis$ where $x=0$ is $y-axis$
The rectangle move from $x=0$ to $x=2$
Consider,
$\begin{array}{rl}& \text{ Area }OABC={\int }_0^2|y|dx\\ \\ & A={\int }_0^2\sqrt{6x+4}dx\\ \\ & A={\int }_0^2(6x+4{)}^{\frac{1}{2}}dx\end{array}$
$\begin{array}{rl}& A=\frac{1}{6}{\left[\frac{(6x+4{)}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^2\\ \\ & A=\frac{2}{18}[(16{)}^{\frac{3}{2}}-(4{)}^{\frac{3}{2}}]\end{array}$
$\begin{array}{rl}& A=\frac{2}{18}[4^3-2^3]\\ \\ & A=\frac{2}{18}[64-8]\\ \\ & A=\frac{2}{18}[56]\\ \\ & A=\frac{56}{9}sq\cdot \text{ unit }\end{array}$

Areas of Bounded Region exercise 20.1 question 9

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

$\begin{array}{rl}& y^2+1=x,x\le 2\\ \\ & y^2+1=x\Rightarrow y=\sqrt{x-1}\\ \\ & \text{ Area of }ABCA=2\times \text{ Area of }ABDA\\ \\ & A=2{\int }_1^2|y|dx\end{array}$
$\begin{array}{rl}& A=2{\int }_1^{2}ydx[y>0\Rightarrow |y|=y]\\ \\ & A=2{\int }_1^2\sqrt{x-1}dx\end{array}$
$\begin{array}{rl}& A=2{\left[\frac{(x-1{)}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_1^2\\ \\ & A=\frac{4}{3}[1^{\frac{3}{2}}-0]\\ \\ & A=\frac{4}{3}sq\cdot \text{ unit }\end{array}$

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$\frac{x^2}{4}+\frac{y^2}{9}=1$and evaluate area of region under curve and above$x-axis$
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 $x-axis$
$\begin{array}{rl}& =2{\int }_0^2|y|\mathrm{dx}\\ \\ & A=2{\int }_0^{2}y\mathrm{dx}\\ \\ & A=2{\int }_0^2\frac{3}{2}\sqrt{4-x^2}dx\end{array}$
$\begin{array}{rl}& A=3{\int }_0^2\sqrt{4-x^2}dx\\ \\ & A=3{[\frac{1}{2}x\sqrt{4-x^2}+\frac{1}{2}4{\sin }^{-1}\frac{x}{2}]}_0^2\end{array}$
$\begin{array}{rl}& A=3[0+\frac{1}{2}\times 4{\sin }^{-1}1]\\ \\ & A=3\times \frac{1}{2}\times 4\times \frac{\pi }{2}\\ \\ & A=3\pi \text{ sq.units }\end{array}$

Areas of Bounded Region exercise 20.1 question 11

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

We have,
$9x^2+4y^2=36$ .........(i)
$\begin{array}{rl}& 4y^2=36-9x^2\\ \\ & y^2=\frac{9}{4}(4-x^2)\\ \\ & y=\frac{3}{2}\sqrt{4-x^2}\end{array}$ ...........(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
$\begin{array}{rl}& A=\text{ Area of enclosed curve }=4{\int }_0^2|y|dx\\ \\ & A=4{\int }_0^2\frac{3}{2}\sqrt{4-x^2}dx\\ \\ & A=4\times \frac{3}{2}{\int }_0^2\sqrt{4-x^2}dx\end{array}$
$\begin{array}{rl}& A=6{\int }_0^2\sqrt{2^2-x^2}dx\\ \\ & A=6{[\frac{x}{2}\sqrt{2^2-x^2}+\frac{1}{2}{2}^2{\sin }^{-1}\frac{x}{2}]}_0^2\\ \\ & A=6[0+\frac{1}{2}4{\sin }^{-1}1]\end{array}$
$\begin{array}{rl}& A=6[\frac{1}{2}\times 4\left(\frac{\pi }{2}\right)]\\ \\ & A=6\pi \mathrm{sq}\cdot \text{ unit }\\ \\ & [\because \int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+c]\text{ and }[\because {\int }_a^b{x}^{n}dx={\left[\frac{x^{n+1}}{n+1}\right]}_a^b]\end{array}$

Areas of Bounded Region exercise 20.1 question 12

Answer:
$\frac{\pi }{2}sq\cdot unit$
Hint:
$[\because \int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+c]$
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
$\begin{array}{rl}& y=2\sqrt{1-x^2}\\ \\ & \frac{y}{2}=\sqrt{1-x^2}\\ \\ & \frac{y^2}{4}=1-x^2\end{array}$
$\begin{array}{rl}{x}^2+\frac{y^2}{4}& =1\\ \\ \frac{x^2}{1}+\frac{y^2}{4}& =1\end{array}$
Since given equation $\frac{x^2}{1}+\frac{y^2}{4}=1$
$\begin{array}{rl}& A={\int }_0^{1}2\sqrt{1-x^2}dx\\ \\ & A=2{\int }_0^1\sqrt{1-x^2}dx\\ \\ & A=2{[\frac{1}{2}x\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x]}_0^1\end{array}$
$\begin{array}{rl}& A=2[\frac{1}{2}{\sin }^{-1}1]\\ \\ & A={\sin }^{-1}1\\ \\ & A=\frac{\pi }{2}sq\cdot \text{ unit }\end{array}$ $[\because \int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+c]$

Areas of Bounded Region exercise 20.1 question 13

Answer:
$\frac{\pi a^2}{4}sq\cdot \text{ unit }$
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:

$\begin{array}{rl}& y=\sqrt{a^2-x^2}\\ \\ & y^2=a^2-x^2\\ \\ & x^2+y^2=a^2\end{array}$
Required area $=$ area of shaded region
$\begin{array}{rl}& ={\int }_0^{a}ydx\\ \\ & ={\int }_0^a\sqrt{a^2-x^2}dx\\ \\ & ={[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}]}_0^a\end{array}$
$\begin{array}{rl}& =[\frac{a}{2}\sqrt{a^2-a^2}+\frac{a^2}{2}{\sin }^{-1}\frac{a}{a}]-[\frac{0}{2}\sqrt{a^2-0}+\frac{a^2}{2}{\sin }^{-1}\frac{0}{a}]\\ \\ & =\frac{a^2}{2}{\sin }^{-1}1\\ \\ & =\frac{a^2}{2}\times \frac{\pi }{2}\\ \end{array}$
$=\frac{\pi a^2}{4}sq\cdot unit$

Areas of Bounded Region exercise 20.1 question 15

Answer:
$\pi a^{2}sq\cdot \text{ unit }$
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$=(0,0)$
Radius$=a$
Hence, $OA=OB=Radius=a$
$A=(a,0),B=(0,a)$
Area of circle$=4\times$ area of region $OBAO$
$=4{\int }_0^{a}ydx$
We know that
$\begin{array}{rl}& x^2+y^2=a^2\\ \\ & y^2=a^2-x^2\\ \\ & y=\pm \sqrt{a^2-x^2}\end{array}$
Since $AOBA$ lies in first quadrant, value of $y$ is positive
$y=\sqrt{a^2-x^2}$
Now,
Area of circle $=4{\int }_0^a\sqrt{a^2-x^2}dx[\because \int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+c]$
$\begin{array}{rl}& =4{[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}]}_0^a\\ \\ & =4[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{a}{a}]-[\frac{0}{2}\sqrt{a^2-0}+\frac{0^2}{2}{\sin }^{-1}\frac{0}{a}]\end{array}$
$\begin{array}{rl}& =4{[0+\frac{a^2}{2}{\sin }^{-1}1]}_0^a\\ \\ & =4\times \frac{a^2}{2}\times \frac{\pi }{2}\\ \\ & =\pi a^2\text{ sq.units }\end{array}$

Areas of Bounded Region exercise 20.1 question 16

Answer:
$\frac{27}{2}sq\cdot \text{ units }$
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 $(-2,2)$ and $x=3$ at $(3,5)$
$y=0$ is $x-axis$

$y=1+|x+1|$

$\{\begin{array}{cc}1-(x+1)& x\le -1\\ 1+(x+1)& x\ge 1\\ -x& x\le -1\\ x+2& x\ge 1\end{array}$
Let required area be $A$ since limits on $x$ are given we use horizontal strip to find area
$\begin{array}{rl}& A={\int }_{-2}^3|y|dx\\ \\ & A={\int }_{-2}^{-1}|y|dx+{\int }_{-1}^3|y|dx\\ \\ & A={\int }_{-2}^{-1}-xdx+{\int }_{-1}^3(x+2)dx\end{array}$
$\begin{array}{rl}& A=-{\left[\frac{x^2}{2}\right]}_{-2}^{-1}+{[\frac{x^2}{2}+2x]}_{-1}^3\\ \\ & A=-[\frac{1}{2}-\frac{4}{2}]+[\frac{9}{2}+6-\frac{1}{2}+2]\\ \\ & A=\frac{3}{2}+8+\frac{8}{2}\end{array}$
$\begin{array}{rl}& A=\frac{3}{2}+8+4\\ \\ & A=\frac{27}{2}sq\cdot unit\end{array}[\because {\int }_a^b{x}^{n}dx={\left[\frac{x^{n+1}}{n+1}\right]}_a^b]$

Areas of Bounded Region exercise 20.1 question 17

Answer:
$\frac{9}{2}sq\cdot unit$
Hint:
Break the limit and find the value of integral.
${\int }_0^1-(x-5)dx$
Given:
Sketch the graph $y=|x-5|$
Evaluate${\int }_0^1|x-5|dx$. What the value of integral represents on graph
Solution:

$\begin{array}{rl}& {\int }_0^1|x-5|dx={\int }_0^1-(x-5)dx\\ \\ & ={[\frac{-x^2}{2}+5x]}_0^1\\ \\ & =[\frac{-1}{2}+5]\end{array}$
$=\frac{9}{2}sq\cdot unit$ $[\because {\int }_a^b{x}^{n}dx={\left[\frac{x^{n+1}}{n+1}\right]}_a^b]$

Areas of Bounded Region exercise 20.1 question 18

Answer:
$9sq\cdot \text{ units }$
Hint:
Break the limit and find integral
Given:
_$y=|x+3|$ and evaluate _{${\int }_{-6}^0|x+3|dx$} What represent of graph
Solution:

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

Areas of Bounded Region exercise 20.1 question 19

Answer:
$9sq\cdot units$
Hint:
Find integral ${\int }_{-4}^2|x+1|dx$
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,
$y=|x+1|$
$=\{\begin{array}{c}(x+1)\text{ for all }x>-1\\ -(x+1)\text{ for all }x<-1\end{array}$
Integral represents the area enclosed between $x=-4$ and $x=2.$
$\begin{array}{rl}A& ={\int }_{-4}^2|y|dx\\ \\ A& ={\int }_{-4}^{-1}|y|dx+{\int }_{-1}^2|y|dx\\ \\ A& ={\int }_{-4}^{-1}-(x+1)dx+{\int }_{-1}^2(x+1)\end{array}$
$\begin{array}{rl}& A=-{[\frac{x^2}{2}+x]}_{-4}^{-1}+{[\frac{x^2}{2}+x]}_{-1}^2\\ \\ & A=-[\frac{1}{2}-1-\frac{16}{2}+4]+[\frac{4}{2}+2-\frac{1}{2}+1]\end{array}$
$\begin{array}{rl}& A=-[3-\frac{15}{2}]+[5-\frac{1}{2}]\\ \\ & A=-3+\frac{15}{2}+5-\frac{1}{2}\\ \\ & A=\frac{-6+15+10-1}{2}\end{array}$
$A=9sq\cdot units$