Question

Match the following definite integrals in column I with their corresponding values in column II and choose the correct option from the codes given below.

Column I		Column II
A	$\int\limits_{-5}^5|x+2| d x$	1	5
B	$\int\limits_2^8|x-5| d x$	2	$\frac{16 \sqrt{2}}{15}$
C	$\int\limits_0^2 x \sqrt{2-x} d x$	3	29
D	$\int\limits_0^4|x-1| d x$	4	9

• A. A - (2), B - (3), C - (1), D - (4)
• B. A - (3), B - (1), C - (2), D - (4)
• C. A - (3), B - (4), C - (2), D - (1)
• D. A - (2), B - (1), C - (3), D - (4)

Answer 1

Answer: (C)

A. Let $I=\int\limits_{-5}^5|x+2| d x$
It can be seen that $(x+2) \leq 0$ on $[-5,-2]$ and $(x+2) \geq 0$ on $[-2,5]$.
$\therefore I =\int\limits_{-5}^{-2}-(x+2) d x+\int\limits_{-2}^5(x+2) d x $
$\Rightarrow {\left[\because \int\limits_a^b f(x) d x=\int\limits_a^c f(x) d x+\int\limits_c^b f(x) d x\right] }$
$=-\left[\frac{x^2}{2}+2 x\right]_{-5}^{-2}+\left[\frac{x^2}{2}+2 x\right]_{-2}^5 $
$ =-\left[\frac{(-2)^2}{2}+2(-2)-\frac{(-5)^2}{2}-2(-5)\right] $
$ +\left[\frac{(5)^2}{2}+2(5)-\frac{(-2)^2}{2}-2(-2)\right]$
$=-\left[2-4-\frac{25}{2}+10\right]+\left[\frac{25}{2}+10-2+4\right]$
$=-2+4+\frac{25}{2}-10+\frac{25}{2}+10-2+4=29$
B. Let $I=\int\limits_2^8|x-5| d x$
It can be seen that $(x-5) \leq 0$ on $[2,5]$ and $(x-5) \geq 0$ on $[5,8]$.
$\because \int\limits_a^b f(x) d x =\int\limits_a^c f(x) d x+\int\limits_c^b f(x) d x $
$\therefore I =\int\limits_2^5\{-(x-5)\} d x+\int\limits_5^8(x-5) d x $
$ =\left[5 x-\frac{x^2}{2}\right]_2^5+\left[\frac{x^2}{2}-5 x\right]_5^8 $
$=\left(25-\frac{25}{2}\right)-\left(10-\frac{4}{2}\right)+\left(\frac{64}{2}-40\right) - \left( \frac{25}{2} - 25\right)$
$ =\frac{25}{2}-8-8+\frac{25}{2}=25-16=9$
C. Let $I=\int\limits_0^2 x \sqrt{2-x} d x$...(i)
Also, $ I=\int\limits_0^2(2-x) \sqrt{2-(2-x)} d x$
$ =\left[\because \int\limits_0^a f(x) d x=\int\limits_0^a f(a-x) d x\right] $
$ =\left[\frac{2 x^{(1 / 2)+1}}{(1 / 2)+1}-\frac{x^{(3 / 2)+1}}{(3 / 2)+1}\right]_0^2=\left[\frac{4}{3} x^{3 / 2}-\frac{2}{5} x^{5 / 2}\right]_0^2 $
$=\frac{4}{3} \cdot 2^{3 / 2}-\frac{2}{5} \cdot 2^{5 / 2}-0=\frac{4}{3} 2 \sqrt{2}-\frac{2}{5} 4 \sqrt{2} $
$ =\left(\frac{8}{3}-\frac{8}{5}\right) \sqrt{2}=\left(\frac{40-24}{15}\right) \sqrt{2}=\frac{16 \sqrt{2}}{15}$
D. Let $I=\int\limits_0^4|x-1| d x$
It can be seen that, $(x-1) \leq 0$ when $0 \leq x \leq 1$ and $(x-1) \geq 0$ when $1 \leq x \leq 4$
$\therefore I=\int\limits_0^1|x-1| d x+\int\limits_0^4|x-1| d x$
$ \left[\because \int\limits_A^b f(x) d x=\int\limits_A^c f(x) d x+\int\limits_C^b f(x) d x\right] $
$ =\int\limits_0^1(1-x) d x+\int\limits_1^4(x-1) d x$
$=\left[x-\frac{x^2}{2}\right]_0^1+\left[\frac{x^2}{2}-x\right]_1^4 $
$ =\left(1-\frac{1}{2}\right)-0+\left(\frac{4^2}{2}-4\right)-\left(\frac{1}{2}-1\right)$
$ =\frac{1}{2}+4+\frac{1}{2}=5$