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

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ We start with the integral equation: $\int_{0}^{\infty} k x e^{-2 x} d x = 1$ ‖ ‖ step 2 ⋮ To solve this, we use the integral of the form $\int_{0}^{\infty} x e^{-a x} d x = \frac{1}{a^2}$ ‖ ‖ step 3 ⋮ Here, $a = 2$, so $\int_{0}^{\infty} x e^{-2 x} d x = \frac{1}{2^2} = \frac{1}{4}$ ‖ ‖ step 4 ⋮ Therefore, $k \cdot \frac{1}{4} = 1$ ‖ ‖ step 5 ⋮ Solving for $k$, we get $k = 4$ ‖ ∻Answer∻ ***C*** ∻Key Concept∻ ⚹Integral of $x e^{-ax}$⚹ ∻Explanation∻ ⚹The integral $\int_{0}^{\infty} x e^{-a x} d x$ evaluates to $\frac{1}{a^2}$. This property is used to solve for $k$ in the given integral equation.⚹◊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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