Question

Match the following

List I	List II
I. ${\int }_{-1}^{1}x\left|x\right|dx$	(a) $\frac{\pi }{2}$
II. ${\int }_0^{\frac{\pi }{2}}(1+\log \left(\frac{4+3\sin x}{4+3\cos x}\right))dx$	(b) ${\int }_0^{\frac{a}{2}}f\left(x\right)dx$
III. ${\int }_0^{a}f\left(x\right)dx$	(c) ${\int }_0^a\left[f\right(x)+f(-x\left)\right]dx$
IV. ${\int }_{-a}^{a}f\left(x\right)dx$	(d) $0$
	(e) ${\int }_0^{a}f(a-x)dx$

• A. I-d; II-a; III-e; IV-c
• B. I-d; II-a; III-c; IV-b
• C. I-d; II-c; III-a; IV-e
• D. I-a; II-d; III-b; IV-c

Answer 1

The correct option is A I-d; II-a; III-e; IV-c

I. ${\int }_{-1}^{1}x|x|dx$

$\Rightarrow$ ${\int }_{-1}^{0}x(-x).dx+{\int }_0^{1}x\left(x\right)dx$

$\Rightarrow$ $[\frac{-x^3}{3}{]}_{-1}^0+[\frac{x^3}{3}{]}_0^1$

$\Rightarrow$ $\frac{-1}{3}+\frac{1}{3}=0$

II. $I={\int }_0^{\pi /2}(1+log(\frac{4+3sinx}{4+3cosx}\left)\right)dx$----(1)

Using the property that, ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$,

$\Rightarrow$ $I={\int }_0^{\pi /2}(1+log(\frac{4+3cosx}{4+sinx}\left)\right)dx$

$\Rightarrow$ $I={\int }_0^{\pi /2}(1-log(\frac{4+3sinx}{4+cosx}\left)\right)dx$----(2)

Adding (1) and (2),

$\Rightarrow$ $2I={\int }_0^{\pi /2}2dx$

$\Rightarrow$ $I=[x{]}_0^{\pi /2}$

$\Rightarrow$ $I=\pi /2$

III. ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$ (it is a standard property of definite integration)

IV. ${\int }_{-a}^{a}f\left(x\right)dx={\int }_0^a\left(f\right(x)+f(-x\left)\right)dx$ (standard property of definite integration)