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Chapter 7: Problem 66

Use integration by parts to evaluate the following integrals. $$\int_{0}^{\infty} x e^{-x} d x$$

Short Answer

Expert verified
Based on the step-by-step solution provided, answer the following question: Question: Evaluate the improper integral \(\int_{0}^{\infty} xe^{-x} dx\). Answer: The value of the given integral is 1.

Step by step solution

01

Choose u and dv

First, we must decide which part of the function should be u and which should be dv. In this case, let: $$u = x$$ $$dv = e^{-x} dx$$
02

Compute du and v

Now let's compute the derivatives and integrals we need: $$ \begin{aligned} du &= dx \\ v &= \int e^{-x} dx \end{aligned} $$ To find the integral of e^{-x}, we can use substitution. Let's say \(t = -x\), then \(dt = -dx\). So, we have: $$ \begin{aligned} v &= \int e^{-x} dx \\ &= -\int e^t dt \\ &= -e^t \\ &= -e^{-x} \end{aligned} $$ Now, we have u, du, v, and dv: $$ \begin{aligned} u &= x \\ du &= dx \\ v &= -e^{-x} \\ dv &= e^{-x} dx \end{aligned} $$
03

Apply integration by parts

Now we will apply the integration by parts formula: $$ \int u dv = uv - \int v du $$ Substitute the values of u, dv, v, and du we found in Steps 1 and 2: $$ \int x e^{-x} dx = (-x e^{-x}) - \int (-e^{-x}) dx $$
04

Evaluate the remaining integral

Now we only need to compute the integral of -e^{-x}. As we did before, we can use substitution to compute this integral: $$ \begin{aligned} -\int e^{-x} dx &= -\int e^t dt \\ &= -(-e^t) \\ &= e^t \\ &= e^{-x} \end{aligned} $$ Now we can substitute this result back into the integration by parts equation: $$ \int x e^{-x} dx = (-x e^{-x}) - e^{-x} + C $$
05

Evaluate the integral with limits

Now let's evaluate the integral using the given limits \(0\) to \(\infty\) by taking the limit of the antiderivative we found as x approaches infinity: $$ \begin{aligned} \lim_{b\to\infty} \left[(-b e^{-b}) - e^{-b}\right] - \left[(-0 e^{-0}) - e^{-0}\right] \\ \end{aligned} $$ The first term \((-b e^{-b})\) approaches 0 as b approaches infinity, and \(e^{-b}\) also approaches 0 as b approaches infinity. Therefore, the value of the integral is: $$ \begin{aligned} -\left[0 - 1\right] = 1 \end{aligned} $$ So, the value of the given integral is 1.

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