your last submission is used for your score. 5. -/6.25 points details my notes evaluate the integral. ∫₁⁴ √t…

Explanation:

Step1: Apply integration - by - parts formula

The integration - by - parts formula is $\int_{a}^{b}u\mathrm{d}v=uv|{a}^{b}-\int{a}^{b}v\mathrm{d}u$. Let $u = \ln(t)$ and $\mathrm{d}v=\sqrt{t}\mathrm{d}t$. Then $\mathrm{d}u=\frac{1}{t}\mathrm{d}t$ and $v=\frac{2}{3}t^{\frac{3}{2}}$.

Step2: Calculate $uv|{a}^{b}$ and $\int{a}^{b}v\mathrm{d}u$

$uv|{1}^{4}=\frac{2}{3}t^{\frac{3}{2}}\ln(t)\big|{1}^{4}=\frac{2}{3}\times4^{\frac{3}{2}}\ln(4)-\frac{2}{3}\times1^{\frac{3}{2}}\ln(1)=\frac{2}{3}\times8\ln(4)-0=\frac{16}{3}\ln(4)$. $\int_{1}^{4}v\mathrm{d}u=\int_{1}^{4}\frac{2}{3}t^{\frac{3}{2}}\times\frac{1}{t}\mathrm{d}t=\frac{2}{3}\int_{1}^{4}t^{\frac{1}{2}}\mathrm{d}t$.

Step3: Evaluate $\frac{2}{3}\int_{1}^{4}t^{\frac{1}{2}}\mathrm{d}t$

$\frac{2}{3}\int_{1}^{4}t^{\frac{1}{2}}\mathrm{d}t=\frac{2}{3}\times\frac{2}{3}t^{\frac{3}{2}}\big|_{1}^{4}=\frac{4}{9}(4^{\frac{3}{2}} - 1^{\frac{3}{2}})=\frac{4}{9}(8 - 1)=\frac{28}{9}$.

Step4: Calculate the original integral

$\int_{1}^{4}\sqrt{t}\ln(t)\mathrm{d}t=\frac{16}{3}\ln(4)-\frac{28}{9}$.

Answer:

$\frac{16}{3}\ln(4)-\frac{28}{9}$