Question

\(\int_0^1 {y{{\left( {{y^2} + 1} \right)}^5}dy} \)

Short answer

Integrate the function,

\(\begin{aligned}{c}\int_0^1 {y{{\left( {{y^2} + 1} \right)}^5}dy} {\rm{ }} &= \frac{1}{2}\left. {\left( {\frac{{{u^6}}}{6}} \right)} \right|_1^2\\ &= \frac{1}{{12}}\left( {{2^6} - {1^6}} \right)\\ &= \frac{{21}}{4}\end{aligned}\)

Step-by-step solution

Use the substitution rule for integration

Let \(u = {y^2} + 1\)

Then,\(du = 2ydy\)

\( \Rightarrow dy = \frac{{du}}{{2y}}\)

New limits of integration for \(u\)

\(u = \left\{ \begin{array}{l}{\rm{2 , if }}x = 1,\\{\rm{1 , if }}x = 0\end{array} \right.\)

Substitute \(u\) to the function \(\int_0^1 {y{{\left( {{y^2} + 1} \right)}^5}dy} \)

\(\begin{aligned}{c}\int_0^1 {y{{\left( {{y^2} + 1} \right)}^5}dy} &= \int_1^2 {\frac{{{u^5}}}{2}du} \\ &= \frac{1}{2}\int_1^2 {{u^5}du} \end{aligned}\)

Use the integral formula \(\int_1^2 {{x^n}dx}  = \frac{{{x^{n + 1}}}}{{n + 1}} + C\)

\(\left. { = \frac{1}{2}\left( {\frac{{{u^6}}}{6}} \right)} \right|_1^2\)

Use the integral formula \(\int_a^b {f\left( x \right)dx}  = F\left( b \right) - F\left( a \right)\)

\(\begin{aligned}{c} &= \frac{1}{{12}}\left( {{2^6} - {1^6}} \right)\\ &= \frac{{63}}{{12}}\\ &= \frac{{21}}{4}\end{aligned}\)

Therefore, the value is \(\frac{{21}}{4}\).