evaluate the integral.\n int_{1}^{4} sqrt{y} ln(y) dy

Explanation:

Step1: Use integration - by - parts

Let $u = \ln(y)$ and $dv=\sqrt{y}dy$. Then $du=\frac{1}{y}dy$ and $v=\frac{2}{3}y^{\frac{3}{2}}$. By the integration - by - parts formula $\int_{a}^{b}u;dv=uv|{a}^{b}-\int{a}^{b}v;du$, we have: [ \begin{align*} \int_{1}^{4}\sqrt{y}\ln(y)dy&=\left[\frac{2}{3}y^{\frac{3}{2}}\ln(y)\right]{1}^{4}-\int{1}^{4}\frac{2}{3}y^{\frac{3}{2}}\cdot\frac{1}{y}dy\ &=\left[\frac{2}{3}y^{\frac{3}{2}}\ln(y)\right]{1}^{4}-\frac{2}{3}\int{1}^{4}y^{\frac{1}{2}}dy \end{align*} ]

Step2: Evaluate $\left[\frac{2}{3}y^{\frac{3}{2}}\ln(y)\right]_{1}^{4}$

[ \begin{align*} \left[\frac{2}{3}y^{\frac{3}{2}}\ln(y)\right]_{1}^{4}&=\frac{2}{3}(4)^{\frac{3}{2}}\ln(4)-\frac{2}{3}(1)^{\frac{3}{2}}\ln(1)\ &=\frac{2}{3}(8)\ln(4)-\frac{2}{3}(1)(0)\ &=\frac{16}{3}\ln(4) \end{align*} ]

Step3: Evaluate $\frac{2}{3}\int_{1}^{4}y^{\frac{1}{2}}dy$

[ \begin{align*} \frac{2}{3}\int_{1}^{4}y^{\frac{1}{2}}dy&=\frac{2}{3}\left[\frac{2}{3}y^{\frac{3}{2}}\right]{1}^{4}\ &=\frac{4}{9}\left[y^{\frac{3}{2}}\right]{1}^{4}\ &=\frac{4}{9}(4^{\frac{3}{2}} - 1^{\frac{3}{2}})\ &=\frac{4}{9}(8 - 1)\ &=\frac{28}{9} \end{align*} ]

Step4: Calculate the final result

[ \begin{align*} \int_{1}^{4}\sqrt{y}\ln(y)dy&=\frac{16}{3}\ln(4)-\frac{28}{9}\ &=\frac{16}{3}(2\ln(2))-\frac{28}{9}\ &=\frac{32}{3}\ln(2)-\frac{28}{9} \end{align*} ]

Answer:

$\frac{32}{3}\ln(2)-\frac{28}{9}$