evaluate the definite integral.\n int_{1}^{3}\frac{8(ln x)^{3}}{x}dx \n int_{1}^{3}\frac{8(ln…

Explanation:

Step1: Use substitution

Let $u = \ln x$, then $du=\frac{1}{x}dx$. When $x = 1$, $u=\ln(1)=0$; when $x = 3$, $u=\ln(3)$. The integral $\int_{1}^{3}\frac{8(\ln x)^{3}}{x}dx$ becomes $8\int_{0}^{\ln(3)}u^{3}du$.

Step2: Integrate $u^{3}$

The antiderivative of $u^{3}$ is $\frac{u^{4}}{4}$. So $8\int_{0}^{\ln(3)}u^{3}du=8\times\left[\frac{u^{4}}{4}\right]_{0}^{\ln(3)}$.

Step3: Evaluate the definite - integral

$8\times\left[\frac{u^{4}}{4}\right]{0}^{\ln(3)} = 2u^{4}\big|{0}^{\ln(3)}=2(\ln(3))^{4}-2(0)^{4}=2(\ln(3))^{4}$. Using a calculator, $2(\ln(3))^{4}\approx2\times(1.0986)^{4}\approx2\times1.4571\approx2.914$.

Answer:

$2.914$