evaluate the definite integral below. ∫-2,-1 6xe^(-x^2 + 1) dx

Explanation:

Step1: Use substitution

Let $u=-x^{2}+1$, then $du=-2xdx$, and $xdx =-\frac{1}{2}du$. When $x = - 2$, $u=-(-2)^{2}+1=-3$. When $x=-1$, $u=-(-1)^{2}+1 = 0$.

Step2: Rewrite the integral

The original integral $\int_{-2}^{-1}6xe^{-x^{2}+1}dx$ becomes $\int_{-3}^{0}6\times(-\frac{1}{2})e^{u}du=-3\int_{-3}^{0}e^{u}du$.

Step3: Integrate $e^{u}$

The antiderivative of $e^{u}$ is $e^{u}$. So, $-3\int_{-3}^{0}e^{u}du=-3\left[e^{u}\right]_{-3}^{0}$.

Step4: Evaluate the definite - integral

$-3\left(e^{0}-e^{-3}\right)=-3\left(1 - \frac{1}{e^{3}}\right)=\frac{3}{e^{3}}-3$.

Answer:

$\frac{3}{e^{3}}-3$