Question

The value of $\frac{29{\int }_0^1{(1-x^4)}^{7}dx}{10{\int }_0^1{(1-x^4)}^{6}dx}$ is

• A. $\frac{29}{28}$
• B. $\frac{28}{29}$
• C. $\frac{14}{5}$
• D. $\frac{5}{14}$

Answer 1

The correct option is C $\frac{14}{5}$

Let, $I={\int }_0^1{(1-x^4)}^7\cdot 1dx$
Applying integration by parts
$={\left[x{(1-x^4)}^7\right]}_0^1+7\times 4{\int }_0^{1}x{(1-x^4)}^6{x}^{3}dx=0-28{\int }_0^1(1-x^4-1){(1-x^4)}^{6}dx=-28{\int }_0^1{(1-x^4)}^{7}dx+28{\int }_0^1{(1-x^4)}^{6}dx=-28I+28{\int }_0^1{(1-x^4)}^{6}dx29I=28{\int }_0^1{(1-x^4)}^{6}dx\Rightarrow \frac{29{\int }_0^1{(1-x^4)}^{7}dx}{10{\int }_0^1{(1-x^4)}^{6}dx}=\frac{28}{10}=\frac{14}{5}$