Question

evaluate the definite integral.
\\(\int_{0}^{1}x^{8}(3 + 2x^{9})^{4}dx\\)

Explanation:

Step1: Use substitution method

Let $u = 3+2x^{9}$. Then $du=18x^{8}dx$, and $x^{8}dx=\frac{1}{18}du$. When $x = 0$, $u=3 + 2\times0^{9}=3$. When $x = 1$, $u=3+2\times1^{9}=5$.

Step2: Rewrite the integral

The original integral $\int_{0}^{1}x^{8}(3 + 2x^{9})^{4}dx$ becomes $\frac{1}{18}\int_{3}^{5}u^{4}du$.

Step3: Integrate $u^{4}$

Using the power - rule for integration $\int u^{n}du=\frac{u^{n + 1}}{n+1}+C$ ($n
eq - 1$), we have $\frac{1}{18}\times\frac{u^{5}}{5}\big|_{3}^{5}$.

Step4: Evaluate the definite integral

$\frac{1}{90}(u^{5})\big|_{3}^{5}=\frac{1}{90}(5^{5}-3^{5})=\frac{1}{90}(3125 - 243)=\frac{1}{90}\times2882=\frac{1441}{45}$.

Answer:

$\frac{1441}{45}$