Question

Matrix -Match type

Match the elements of List I with the elements of List II and choose the correct match:

List-I	List-II
A) ${\int }_0^{\infty }{e}^{-4x}\sin 5xdx$	P. $3$
B) ${\int }_2^8\frac{\left[x^2\right]dx}{[x^2-20x+100]+\left[x^2\right]}$ Where $[.]$ is greatest integer function $\le x$	Q. $\frac{5}{41}$
C) ${\int }_0^{\frac{\pi }{2}}[x^n\cos x+n{x}^{n-1}\sin x]dx$ Where $n\in N$	R. $120$
D) ${\int }_0^{\infty }{x}^5{e}^{-x}dx$	S. ${\left(\frac{\pi }{2}\right)}^n$

• A. $A-Q,B-P,C-S,D-R$
• B. $A-P,B-Q,C-R,D-S$
• C. $A-Q,B-P,C-R,D-S$
• D. $A-P,B-Q,C-S,D-R$

Answer 1

Answer: (A)

$A-Q,B-P,C-S,D-R$

A) $I={\int }_0^{\infty }{e}^{-4x}\sin 5xdx$

Applying integration by parts,

$={[\sin 5x.\frac{e^{-4x}}{-4}]}_0^{\infty }-{\int }_0^{\infty }5\cos 5x.\frac{e^{-4x}}{-4}dx$

$=0+\frac{5}{4}{\int }_0^{\infty }{e}^{-4x}\cos 5xdx$

Again applying integration by parts,

$I=\frac{5}{4}{\left[\cos 5x\frac{e^{-4x}}{-4}\right]}_0^{\infty }-\frac{5}{4}{\int }_0^{\infty }-5\sin 5x\frac{e^{-4x}}{-4}dx$

$I=\frac{5}{4}[0-\frac{1}{-4}]-\frac{25}{16}{\int }_0^{\infty }{e}^{-4x}\sin 5xdx$

$I=\frac{5}{16}-\frac{25}{16}I$

$\Rightarrow I=\frac{5}{41}$

B) ${\int }_2^8\frac{\left[x^2\right]dx}{[x^2-20x+100]+\left[x^2\right]}$

$I={\int }_2^8\frac{\left[x^2\right]dx}{\left[\right(x-10{)}^2]+[x^2]}$ .....(1)

$I={\int }_2^8\frac{\left[\right(10-x{)}^2]dx}{\left[x^2\right]+\left[\right(10-x{)}^2]}$ ....(2) $\left({\int }_a^{b}f\right(x)dx={\int }_a^{b}f(a+b-x\left)dx\right)$

Adding (1) and (2),

$2I={\int }_2^{8}1dx$

$\Rightarrow 2I=[x{]}_2^8$

$\Rightarrow I=3$

C) $I={\int }_0^{\frac{\pi }{2}}(x^n\cos x+n{x}^{n-1}\sin x)dx$

Assume, $x^n\sin x=t\Rightarrow x^n\cos x+n{x}^{n-1}\sin xdx=dt$

$I={\int }_0^{{\left(\frac{\pi }{2}\right)}^n}dt$

$I=[t{]}_0^{{\left(\frac{\pi }{2}\right)}^n}$

$I={\left(\frac{\pi }{2}\right)}^n$

D) $I={\int }_0^{\infty }{x}^5{e}^{-x}dx$

$I={\left[x^5\frac{e^{-x}}{-1}\right]}_0^{\infty }-{\int }_0^{\infty }5x^4\frac{e^{-x}}{-1}dx$

$I=0+5{\int }_0^{\infty }{x}^4{e}^{-x}dx$

$I=5{\left[x^4\frac{e^{-x}}{-1}\right]}_0^{\infty }-5{\int }_0^{\infty }4x^3\frac{e^{-x}}{-1}dx$

$I=0+20{\int }_0^{\infty }{x}^3{e}^{-x}dx$

$I=20{\left[x^3\frac{e^{-x}}{-1}\right]}_0^{\infty }-20{\int }_0^{\infty }3x^2\frac{e^{-x}}{-1}dx$

$I=60{\int }_0^{\infty }{x}^2{e}^{-x}dx$

$I=60{\left[x^2\frac{e^{-x}}{-1}\right]}_0^{\infty }-60{\int }_0^{\infty }2x\frac{e^{-x}}{-1}dx$

$I=0+120{\int }_0^{\infty }x{e}^{-x}dx$

$I=120{\left[x\frac{e^{-x}}{-1}\right]}_0^{\infty }-120{\int }_0^{\infty }\frac{e^{-x}}{-1}dx$

$I=120{\int }_0^{\infty }{e}^{-x}dx$

$\Rightarrow I=120$