evaluate the definite integral.\n int_{2}^{8} t^{2} ln(2t)dt

Explanation:

Step1: Use integration - by - parts formula

The integration - by - parts formula is $\int u;dv=uv-\int v;du$. Let $u = \ln(2t)$ and $dv=t^{2}dt$. Then $du=\frac{1}{t}dt$ and $v=\frac{1}{3}t^{3}$.

Step2: Apply the integration - by - parts formula

$\int t^{2}\ln(2t)dt=\frac{1}{3}t^{3}\ln(2t)-\int\frac{1}{3}t^{3}\cdot\frac{1}{t}dt=\frac{1}{3}t^{3}\ln(2t)-\frac{1}{3}\int t^{2}dt$.

Step3: Integrate $t^{2}$

$\int t^{2}dt=\frac{1}{3}t^{3}+C$, so $\int t^{2}\ln(2t)dt=\frac{1}{3}t^{3}\ln(2t)-\frac{1}{9}t^{3}+C$.

Step4: Evaluate the definite integral

$\left[\frac{1}{3}t^{3}\ln(2t)-\frac{1}{9}t^{3}\right]_{2}^{8}=\left(\frac{1}{3}(8)^{3}\ln(16)-\frac{1}{9}(8)^{3}\right)-\left(\frac{1}{3}(2)^{3}\ln(4)-\frac{1}{9}(2)^{3}\right)$. First, simplify the terms: $\frac{1}{3}(8)^{3}\ln(16)=\frac{512}{3}\ln(2^{4})=\frac{2048}{3}\ln(2)$, $\frac{1}{9}(8)^{3}=\frac{512}{9}$, $\frac{1}{3}(2)^{3}\ln(4)=\frac{8}{3}\ln(2^{2})=\frac{16}{3}\ln(2)$, $\frac{1}{9}(2)^{3}=\frac{8}{9}$. Then, $\left(\frac{2048}{3}\ln(2)-\frac{512}{9}\right)-\left(\frac{16}{3}\ln(2)-\frac{8}{9}\right)=\left(\frac{2048}{3}\ln(2)-\frac{16}{3}\ln(2)\right)+\left(-\frac{512}{9}+\frac{8}{9}\right)$. Combining like - terms: $\frac{2048 - 16}{3}\ln(2)-\frac{512 - 8}{9}=\frac{2032}{3}\ln(2)-\frac{504}{9}=\frac{2032}{3}\ln(2)- 56$.

Answer:

$\frac{2032}{3}\ln(2)-56$