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	$\underset{-5}{\overset{5}{\int }}|x+2|dx$	1	5
B	$\underset{2}{\overset{8}{\int }}|x-5|dx$	2	$\frac{16\sqrt{2}}{15}$
C	$\underset{0}{\overset{2}{\int }}x\sqrt{2-x}dx$	3	29
D	$\underset{0}{\overset{4}{\int }}|x-1|dx$	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 $I=\underset{-5}{\overset{5}{\int }}|x+2|dx$
It can be seen that $(x+2)\le 0$ on $[-5,-2]$ and $(x+2)\ge 0$ on $[-2,5]$.
$\therefore I=\underset{-5}{\overset{-2}{\int }}-(x+2)dx+\underset{-2}{\overset{5}{\int }}(x+2)dx$
$\Rightarrow \left[\because \underset{a}{\overset{b}{\int }}f(x)dx=\underset{a}{\overset{c}{\int }}f(x)dx+\underset{c}{\overset{b}{\int }}f(x)dx\right]$
$=-{\left[\frac{x^2}{2}+2x\right]}_{-5}^{-2}+{\left[\frac{x^2}{2}+2x\right]}_{-2}^5$
$=-\left[\frac{(-2{)}^2}{2}+2(-2)-\frac{(-5{)}^2}{2}-2(-5)\right]$
$+\left[\frac{(5{)}^2}{2}+2(5)-\frac{(-2{)}^2}{2}-2(-2)\right]$
$=-\left[2-4-\frac{25}{2}+10\right]+\left[\frac{25}{2}+10-2+4\right]$
$=-2+4+\frac{25}{2}-10+\frac{25}{2}+10-2+4=29$
B. Let $I=\underset{2}{\overset{8}{\int }}|x-5|dx$
It can be seen that $(x-5)\le 0$ on $[2,5]$ and $(x-5)\ge 0$ on $[5,8]$.
$\because \underset{a}{\overset{b}{\int }}f(x)dx=\underset{a}{\overset{c}{\int }}f(x)dx+\underset{c}{\overset{b}{\int }}f(x)dx$
$\therefore I=\underset{2}{\overset{5}{\int }}\{-(x-5)\}dx+\underset{5}{\overset{8}{\int }}(x-5)dx$
$={\left[5x-\frac{x^2}{2}\right]}_2^5+{\left[\frac{x^2}{2}-5x\right]}_5^8$
$=\left(25-\frac{25}{2}\right)-\left(10-\frac{4}{2}\right)+\left(\frac{64}{2}-40\right)-\left(\frac{25}{2}-25\right)$
$=\frac{25}{2}-8-8+\frac{25}{2}=25-16=9$
C. Let $I=\underset{0}{\overset{2}{\int }}x\sqrt{2-x}dx$...(i)
Also, $I=\underset{0}{\overset{2}{\int }}(2-x)\sqrt{2-(2-x)}dx$
$=\left[\because \underset{0}{\overset{a}{\int }}f(x)dx=\underset{0}{\overset{a}{\int }}f(a-x)dx\right]$
$={\left[\frac{2x^{(1/2)+1}}{(1/2)+1}-\frac{x^{(3/2)+1}}{(3/2)+1}\right]}_0^2={\left[\frac{4}{3}{x}^{3/2}-\frac{2}{5}{x}^{5/2}\right]}_0^2$
$=\frac{4}{3}\cdot 2^{3/2}-\frac{2}{5}\cdot 2^{5/2}-0=\frac{4}{3}2\sqrt{2}-\frac{2}{5}4\sqrt{2}$
$=\left(\frac{8}{3}-\frac{8}{5}\right)\sqrt{2}=\left(\frac{40-24}{15}\right)\sqrt{2}=\frac{16\sqrt{2}}{15}$
D. Let $I=\underset{0}{\overset{4}{\int }}|x-1|dx$
It can be seen that, $(x-1)\le 0$ when $0\le x\le 1$ and $(x-1)\ge 0$ when $1\le x\le 4$
$\therefore I=\underset{0}{\overset{1}{\int }}|x-1|dx+\underset{0}{\overset{4}{\int }}|x-1|dx$
$\left[\because \underset{A}{\overset{b}{\int }}f(x)dx=\underset{A}{\overset{c}{\int }}f(x)dx+\underset{C}{\overset{b}{\int }}f(x)dx\right]$
$=\underset{0}{\overset{1}{\int }}(1-x)dx+\underset{1}{\overset{4}{\int }}(x-1)dx$
$={\left[x-\frac{x^2}{2}\right]}_0^1+{\left[\frac{x^2}{2}-x\right]}_1^4$
$=\left(1-\frac{1}{2}\right)-0+\left(\frac{4^2}{2}-4\right)-\left(\frac{1}{2}-1\right)$
$=\frac{1}{2}+4+\frac{1}{2}=5$