Question

The value of ${\int }_0^1{x}^4{(1-x)}^{4}dx$

• A. $\frac{1}{630}$
• B. $\frac{2}{105}$
• C. $0$
• D. $\frac{71}{1545}$

Answer 1

The correct option is A $\frac{1}{630}$

Consider the given integral.

$I={\int }_0^1{x}^4{(1-x)}^{4}dx$

$I={\int }_0^1{x}^4{(1-x)}^2{(1-x)}^{2}dx$

$I={\int }_0^1{x}^4(1+x^2-2x)(1+x^2-2x)dx$

$I={\int }_0^1{x}^4(1+x^2-2x+x^2+x^4-2x^3-2x-2x^3+4x^2)dx$

$I={\int }_0^1{x}^4(x^4-4x^3+6x^2-4x+1)dx$

$I={\int }_0^1(x^8-4x^7+6x^6-4x^5+x^4)dx$

$I={[\frac{x^9}{9}-\frac{4x^8}{8}+\frac{6x^7}{7}-\frac{4x^6}{6}+\frac{x^5}{5}]}_0^1$

$I=[\frac{1^9}{9}-\frac{4\times 1^8}{8}+\frac{6\times 1^7}{7}-\frac{4\times 1^6}{6}+\frac{1^5}{5}-0]$

$I=\frac{1}{9}-\frac{1}{2}+\frac{6}{7}-\frac{2}{3}+\frac{1}{5}$

$I=\frac{210-945+1620-1260+378}{1890}$

$I=\frac{2208-2205}{1890}$

$I=\frac{3}{1890}=\frac{1}{630}$

Hence, this is the answer.