determine the following. ∫4e^(-0.4x) dx ∫4e^(-0.4x) dx = □ (type an exact answer.)

Explanation:

Step1: Use integration - by - substitution

Let $u=-0.4x$, then $du=-0.4dx$ and $dx =-\frac{1}{0.4}du$. The integral $\int4e^{-0.4x}dx$ becomes $4\int e^{u}\left(-\frac{1}{0.4}\right)du$.

Step2: Simplify the integral

$4\int e^{u}\left(-\frac{1}{0.4}\right)du=- \frac{4}{0.4}\int e^{u}du$.

Step3: Integrate $e^{u}$

Since $\int e^{u}du = e^{u}+C$, then $-\frac{4}{0.4}\int e^{u}du=-\frac{4}{0.4}e^{u}+C$.

Step4: Substitute back $u = - 0.4x$

$-\frac{4}{0.4}e^{-0.4x}+C=-10e^{-0.4x}+C$.

Answer:

$-10e^{-0.4x}+C$