3. For what value of $k$, if any, is $\int_{0}^{\infty} k x e^{-2 x} d x=1$ ?
(A) $\frac{1}{4}$
(B) 1
(C) 4
(D) There is no such value of $k$.

Question

3. For what value of $k$, if any, is $\int_{0}^{\infty} k x e^{-2 x} d x=1$ ?
(A) $\frac{1}{4}$
(B) 1
(C) 4
(D) There is no such value of $k$.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Evaluate the improper integral $\int_{0}^{\infty} k x e^{-2 x} dx$ for a constant $k$. ‖ ‖ step 2 ⋮ Use integration by parts with $u = x$ and $dv = k e^{-2 x} dx$. ‖ ‖ step 3 ⋮ Find $du = dx$ and $v = -\frac{k}{2} e^{-2 x}$. ‖ ‖ step 4 ⋮ Apply the integration by parts formula: $\int u dv = uv - \int v du$. ‖ ‖ step 5 ⋮ Calculate the integral: $\left[-\frac{k}{2} x e^{-2 x}\right]_{0}^{\infty} + \int_{0}^{\infty} \frac{k}{2} e^{-2 x} dx$. ‖ ‖ step 6 ⋮ Evaluate the limits and the remaining integral to find the value of $k$ such that the integral equals 1. ‖ ‖ step 7 ⋮ Solve the equation $\frac{k}{4} = 1$ for $k$. ‖ ‖ step 8 ⋮ Find that $k = 4$. ‖ ∻Answer∻ ⚹ (C) 4 ⚹ ∻Key Concept∻ ⚹Integration by parts and evaluation of improper integrals⚹ ∻Explanation∻ ⚹The value of $k$ is found by setting up the integral using integration by parts and solving for $k$ when the integral equals 1.⚹◊What is the value of $k$ in the integral $\int_{0}^{\infty} k x e^{-2x} dx = 1$?⍭ Generate me a similar question◊

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