Question

Match the following:

List-I	List-II
1. Area of region bounded by $y=2x-x^2$ and $x-$axis	a. $\frac{1}{3}$
2. Area of the region $\left\{\right(x,y):x^2\le y\le |x|\}$	b. $\frac{1}{2}$
3. Area bounded by $y=x$ and $y=x^3$	c. $\frac{2}{3}$
4. Area bounded by $y=x|x|$, $x$-axis and $x=-1,x=1$	d. $\frac{4}{3}$

The correct match for $1234$ is

• A. $1-b,2-c.3-d,4-a$
• B. $1-c,2-d,3-a,4-b$
• C. $1-d,2-a,3-b,4-c$
• D. $1-a,2-b,3-c,4-d$

Answer 1

Answer: (C)

$1-d,2-a,3-b,4-c$

1:

$y=2x-x^2$

The values of $x$ where $y$ is $0$ are

$x=0,2$

Now, the area of the region bounded by $y=2x-x^2$ and $x-$axis is

${\int }_0^2(2x-x^2)dx={[x^2-\frac{x^3}{3}]}_0^2$

$=4-\frac{8}{3}=\frac{4}{3}\to d$

2:

$x^2\le y\le |x|$

We have to find the area of the region bounded by $y=x^2$ and $y=|x|$ for $x\in [-1,1]$

${\int }_{-1}^1-(x^2-|x|)dx=-2{\int }_0^1(x^2-x)dx$

$=-2{[\frac{x^3}{3}-\frac{x^2}{2}]}_0^1=\frac{1}{3}\to a$

3:

We have to find the area of the region bounded by $y=x$ and $y=x^3$ for $x\in [-1,1]$

${\int }_{-1}^1(x-x^3)dx=2{\int }_0^1(x-x^3)dx$

$=2{[\frac{x^2}{2}-\frac{x^4}{4}]}_0^1=\frac{1}{4}\to b$

4:

We have to find the area of the region bounded by $y=x|x|$ and $y=0$ for $x\in [-1,1]$

${\int }_{-1}^1\left(x|x|\right)dx=2{\int }_0^1\left(x^2\right)dx$

$=2{\left[\frac{x^3}{3}\right]}_0^1=\frac{2}{3}\to c$

Hence, option C.