use integration by parts to evaluate the definite integral ∫₀⁵ te⁻ᵗ dt.
use integration by parts to evaluate the definite integral ∫₀⁵ te⁻ᵗ dt.
Answer
Explanation:
Step1: Recall integration - by - parts formula
The integration - by - parts formula for definite integrals is $\int_{a}^{b}u;dv=uv|{a}^{b}-\int{a}^{b}v;du$. Let $u = t$ and $dv=e^{-t}dt$.
Step2: Find $du$ and $v$
Differentiate $u = t$ with respect to $t$ to get $du = dt$. Integrate $dv = e^{-t}dt$ with respect to $t$: $\int e^{-t}dt=-e^{-t}+C$, so $v=-e^{-t}$.
Step3: Apply the integration - by - parts formula
$\int_{0}^{5}te^{-t}dt=\left[-te^{-t}\right]{0}^{5}-\int{0}^{5}(-e^{-t})dt$. First, evaluate $\left[-te^{-t}\right]{0}^{5}=-5e^{-5}-(-0\times e^{0})=- \frac{5}{e^{5}}$. Second, evaluate $\int{0}^{5}(-e^{-t})dt=\left[e^{-t}\right]_{0}^{5}=e^{-5}-e^{0}=\frac{1}{e^{5}} - 1$.
Step4: Calculate the final result
$\int_{0}^{5}te^{-t}dt=-\frac{5}{e^{5}}-(\frac{1}{e^{5}} - 1)=-\frac{5}{e^{5}}-\frac{1}{e^{5}} + 1=1-\frac{6}{e^{5}}$.
Answer:
$1-\frac{6}{e^{5}}$