Mathematics

$\int_0^\infty {{x^n}{e^{ - x}}dx}$ (n is +ve integer) is equal to

$n!$

SOLUTION
$\displaystyle \int _{ 0 }^{ \propto }{ { x }^{ n }{ e }^{ -x }dx= } \Gamma \left( n+1 \right)$ $=n!$

$\because gamma function$

Its FREE, you're just one step away

Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

Realted Questions

Q1 Subjective Medium
$\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \dfrac { \sin { ^{ 2 }x } }{ \sin { x } +\cos { x } } dx } =-\dfrac { 1 }{ 2\sqrt { 2 } } \log { \left(-2 \sqrt { 2 } +3\right) }$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\displaystyle\int{\frac{x^2(1-\ln{x})}{\ln^4{x}-x^4}dx}$ is equal to
• A. $\displaystyle\frac{1}{2}\ln{\left(\frac{x}{\ln{x}}\right)}-\frac{1}{4}\ln{(\ln^2{x}-x^2)}+C$
• B. $\displaystyle\frac{1}{4}\ln{\left(\frac{\ln(x)+x}{\ln{x}-x}\right)}-\frac{1}{2}\tan^{-1}{\left(\frac{\ln{x}}{x}\right)}+C$
• C. $\displaystyle\frac{1}{4}\left(\ln{\left(\frac{\ln(x)-x}{\ln{x}+x}\right)}+\tan^{-1}{\left(\frac{\ln{x}}{x}\right)}\right)+C$
• D. $\displaystyle\frac{1}{4}\ln{\left(\frac{\ln(x)-x}{\ln{x}+x}\right)}-\frac{1}{2}\tan^{-1}{\left(\frac{\ln{x}}{x}\right)}+C$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
The value of $\int _{ 0 }^{ { \sin }^{ 2 }x }{ { \sin }^{ -1 }\sqrt { t } dt+ } \int _{ 0 }^{ { \cos }^{ 2 }x }{ { \cos }^{ -1 }\sqrt { t }dt}$ is
• A. $\pi$
• B. $\dfrac { \pi }{ 4 }$
• C. $1$
• D. $\dfrac { \pi }{ 2 }$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate the following integrals:
$\displaystyle \int { \tan ^{ 3 }{ x } \sec ^{ 2 }{ x } } dx$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
Let us consider the integral of the following forms
$f{(x_1,\sqrt{mx^2+nx+p})}^{\tfrac{1}{2}}$
Case I If $m>0$, then put $\sqrt{mx^2+nx+C}=u\pm x\sqrt{m}$
Case II If $p>0$, then put $\sqrt{mx^2+nx+C}=u\pm \sqrt{p}$
Case III If quadratic equation $mx^2+nx+p=0$ has real roots $\alpha$ and $\beta$, then put $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020