evaluate the integral. (remember the constant of integration.) ∫te^(-4t) dt

Explanation:

Step1: Apply integration - by - parts formula

The integration - by - parts formula is $\int u\mathrm{d}v=uv-\int v\mathrm{d}u$. Let $u = t$ and $\mathrm{d}v=e^{-4t}\mathrm{d}t$. Then $\mathrm{d}u=\mathrm{d}t$ and $v=-\frac{1}{4}e^{-4t}$.

Step2: Substitute into the formula

$\int te^{-4t}\mathrm{d}t=t\left(-\frac{1}{4}e^{-4t}\right)-\int\left(-\frac{1}{4}e^{-4t}\right)\mathrm{d}t$.

Step3: Evaluate the remaining integral

$\int\left(-\frac{1}{4}e^{-4t}\right)\mathrm{d}t=\frac{1}{16}e^{-4t}+C$. So $\int te^{-4t}\mathrm{d}t=-\frac{1}{4}te^{-4t}+\frac{1}{16}e^{-4t}+C$.

Answer:

$-\frac{1}{4}te^{-4t}+\frac{1}{16}e^{-4t}+C$