evaluate the definite integral.\n int_{1}^{4}\frac{8(ln x)^{3}}{x}dx \n int_{1}^{4}\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 $x = 4$, $u=\ln(4)$. The integral $\int_{1}^{4}\frac{8(\ln x)^{3}}{x}dx$ becomes $8\int_{0}^{\ln(4)}u^{3}du$.

Step2: Integrate $u^{3}$

The antiderivative of $u^{3}$ is $\frac{u^{4}}{4}$ according to the power - rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). So $8\int_{0}^{\ln(4)}u^{3}du=8\times\left[\frac{u^{4}}{4}\right]_{0}^{\ln(4)}$.

Step3: Evaluate the definite integral

$8\times\left[\frac{u^{4}}{4}\right]{0}^{\ln(4)} = 2\left[u^{4}\right]{0}^{\ln(4)}=2((\ln(4))^{4}-0^{4})=2(\ln(4))^{4}$. Using a calculator, $\ln(4)\approx1.3863$, and $2(\ln(4))^{4}=2\times(1.3863)^{4}\approx2\times3.6356\approx7.271$.

Answer:

$7.271$