evaluate the definite integral below. \n int_{-3}^{-1} left(18x(3x^{2}-2)^{2}\right) dx

Explanation:

Step1: Use substitution

Let $u = 3x^{2}-2$, then $du=6xdx$. When $x=-3$, $u = 3\times(-3)^{2}-2=27 - 2=25$. When $x=-1$, $u=3\times(-1)^{2}-2=3 - 2 = 1$. And $18xdx = 3du$.

Step2: Rewrite the integral

The integral $\int_{-3}^{-1}18x(3x^{2}-2)^{2}dx$ becomes $\int_{25}^{1}3u^{2}du$.

Step3: Integrate

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\int3u^{2}du=3\times\frac{u^{3}}{3}=u^{3}+C$.

Step4: Evaluate the definite integral

$[u^{3}]_{25}^{1}=1^{3}-25^{3}=1 - 15625=-15624$.

Answer:

$-15624$