use integration by parts to evaluate the definite integral ∫₀² te⁻ᵗ dt.

Explanation:

Step1: Recall integration - by - parts formula

The integration - by - parts formula for definite integrals is $\int_{a}^{b}u\mathrm{d}v=uv|{a}^{b}-\int{a}^{b}v\mathrm{d}u$. Let $u = t$ and $\mathrm{d}v=e^{-t}\mathrm{d}t$. Then $\mathrm{d}u=\mathrm{d}t$ and $v=-e^{-t}$.

Step2: Apply the formula

$\int_{0}^{2}te^{-t}\mathrm{d}t=\left[-te^{-t}\right]{0}^{2}-\int{0}^{2}-e^{-t}\mathrm{d}t$. First, evaluate $\left[-te^{-t}\right]{0}^{2}$: When $t = 2$, $-te^{-t}=-2e^{-2}$; when $t = 0$, $-te^{-t}=0$. So $\left[-te^{-t}\right]{0}^{2}=-2e^{-2}-0=- \frac{2}{e^{2}}$. Second, evaluate $\int_{0}^{2}-e^{-t}\mathrm{d}t$. We know that $\int - e^{-t}\mathrm{d}t=e^{-t}+C$. Then $\int_{0}^{2}-e^{-t}\mathrm{d}t=\left[e^{-t}\right]_{0}^{2}=e^{-2}-e^{0}=\frac{1}{e^{2}} - 1$.

Step3: Calculate the result

$\int_{0}^{2}te^{-t}\mathrm{d}t=-\frac{2}{e^{2}}-\left(\frac{1}{e^{2}} - 1\right)=-\frac{2}{e^{2}}-\frac{1}{e^{2}}+1=1-\frac{3}{e^{2}}$.

Answer:

$1-\frac{3}{e^{2}}$