evaluate.\n int_{0}^{b} e^{-7x + 5} dx \n int_{0}^{b} e^{-7x + 5} dx=\text{ (type an exact answer.)}

Explanation:

Step1: Use substitution

Let $u=-7x + 5$, then $du=-7dx$ and $dx=-\frac{1}{7}du$. When $x = 0$, $u = 5$; when $x = b$, $u=-7b + 5$.

Step2: Rewrite the integral

$\int_{0}^{b}e^{-7x + 5}dx=-\frac{1}{7}\int_{5}^{-7b + 5}e^{u}du$.

Step3: Integrate $e^u$

The antiderivative of $e^{u}$ is $e^{u}$. So $-\frac{1}{7}\int_{5}^{-7b + 5}e^{u}du=-\frac{1}{7}[e^{u}]_{5}^{-7b + 5}$.

Step4: Evaluate the definite - integral

$-\frac{1}{7}(e^{-7b + 5}-e^{5})=\frac{e^{5}-e^{-7b + 5}}{7}$.

Answer:

$\frac{e^{5}-e^{-7b + 5}}{7}$