evaluate.\n int_{3}^{4}x(x^{2}-9)^{2}dx \n int_{3}^{4}x(x^{2}-9)^{2}dx=square \n(type an exact answer.)

Explanation:

Step1: Use substitution

Let $u = x^{2}-9$, then $du = 2x\ dx$, and $x\ dx=\frac{1}{2}du$. When $x = 3$, $u=3^{2}-9 = 0$; when $x = 4$, $u=4^{2}-9=7$.

Step2: Rewrite the integral

The integral $\int_{3}^{4}x(x^{2}-9)^{2}dx$ becomes $\frac{1}{2}\int_{0}^{7}u^{2}du$.

Step3: Integrate $u^{2}$

Using the power - rule for integration $\int u^{n}du=\frac{u^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\frac{1}{2}\times\frac{u^{3}}{3}\big|_{0}^{7}$.

Step4: Evaluate the definite integral

$\frac{1}{6}(7^{3}-0^{3})=\frac{343}{6}$.

Answer:

$\frac{343}{6}$