Mathematics

# $\displaystyle \int_{0}^{1}x\left ( 1-x \right )^{4}dx= \frac{1}{C}$, then $C=?$

30

##### SOLUTION
Let $\displaystyle I=\int _{ 0 }^{ 1 } x\left( 1-x \right) ^{ 4 }dx$
Using $\int _{ a }^{ b }{ f\left( x \right) dx } =\int _{ a }^{ b }{ f\left( a+b-x \right) dx }$
$\displaystyle \therefore I=\int _{ 0 }^{ 1 } \left( 1-x \right) x^{ 4 }dx=\int _{ 0 }^{ 1 } \left( x^{ 4 }-x^{ 5 } \right) dx=\frac { 1 }{ 5 } -\frac { 1 }{ 6 } =\frac { 1 }{ 30 }$

Its FREE, you're just one step away

One Word Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

#### Realted Questions

Q1 Subjective Medium
Evaluate
$\displaystyle\int { \dfrac { { x }^{ 3 }+{ 4x }^{ 2 }-7x+5 }{ x+2 } } dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
If $f(x)$ is an even function, then $\displaystyle \int_{0}^{x}f(t)\, dt$ is
• A. Even function
• B. Neither even nor odd
• C. None of the above
• D. Odd function

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
Prove that $\displaystyle \int _{\frac{\pi}{3}}^{\frac{\pi}{6}} \dfrac{dx}{1+\sqrt{\cot\, x}} =\int _{\frac{\pi}{3}}^{\frac{\pi}{6}} \dfrac{dx}{1+\sqrt{\tan\, x}}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
$\displaystyle \int _{ 2 }^{ 3 }{ \left( 1+2x \right) } dx$

$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$