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Chapter 11: Problem 899

\(\int(x-1) e^{-x} d x=\square+c\)

Short Answer

Expert verified
The integral of \((x-1)e^{-x} dx\) can be found using integration by parts, with \(u = x-1\) and \(dv = e^{-x} dx\). We then compute \(du = dx\) and \(v = -e^{-x}\) and apply the integration by parts formula, resulting in the final answer: \(-e^{-x}(x-1) - e^{-x} + c\), where \(c\) is the constant of integration.

Step by step solution

01

Understanding the Integration by Parts Formula

The integration by parts formula is given by \(\int u \ dv = u \cdot v - \int v \ du\), where \(u\) and \(dv\) are parts of the integrand that are selected in a way that makes the right-hand side simpler than the original integral. For the given problem, we need to carefully choose \(u\) and \(dv\) to simplify our integral.
02

Choose \(u\) and \(dv\) Appropriately

In this instance, let \(u = x - 1\) and \(dv = e^{-x} dx\). The reason behind this choice is that the derivative of \(x - 1\) will be simplified (i.e., the exponent of \(x\) will reduce), and the integral of \(e^{-x} dx\) is straightforward to find.
03

Compute \(du\) and \(v\)

After choosing \(u\) and \(dv\), find \(du\) (the derivative of \(u\)) and \(v\) (which is the integral of \(dv\)). This results in \(du = dx\) and \(v = -e^{-x}\).
04

Apply the Integration by Parts Formula

Substituting our values of \(u\), \(dv\), \(du\) and \(v\) into the integration by parts formula, we get \(\int (x - 1)e^{-x} dx = (x - 1) (-e^{-x}) - \int -e^{-x} dx\).
05

Simplify the Integral

We then simplify the expression to get: \(-e^{-x}(x - 1) + \int e^{-x} dx\).
06

Evaluate Remaining Integral

The integral \(\int e^{-x} dx\) is easy to compute; it is \(-e^{-x}\), because the derivative of \(-e^{-x}\) gives \(e^{-x}\).
07

Final Result

Substituting back in, we get: \(-e^{-x}(x - 1) - e^{-x} + c\), where \(c\) is the constant of integration. Thus, the integral of \((x - 1)e^{-x}\) is \(-e^{-x}(x - 1) - e^{-x} + c\).

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Most popular questions from this chapter

\(\int \mathrm{e}^{2 \mathrm{x}+\log \mathrm{x}} \mathrm{dx}=\) (a) \((1 / 4)(2 \mathrm{x}-1) \mathrm{e}^{2 \mathrm{x}}\) (b) \((1 / 2)(2 \mathrm{x}-1) \mathrm{e}^{2 \mathrm{x}}\) (c) \((1 / 4)(2 \mathrm{x}+1) \mathrm{e}^{2 \mathrm{x}}\) (d) \((1 / 4)(2 \mathrm{x}+1) \mathrm{e}^{2 \mathrm{x}}\)

\(\int[(x-\sin x) /(1-\cos x)] d x=\) \(+c\) (a) \(x \tan (x / 2)\) (b) \(-x \cot (x / 2)\) (c) \(\cot (\mathrm{x} / 2)\) (d) \(-\cot (x / 2)\)

If \(\int\left[\left(x^{4}+1\right) /\left(x^{6}+1\right)\right] d x=\tan ^{-1} x+P \tan ^{-1} x^{3}+c\) then \(P=\) (a) 3 (b) \((1 / 3)\) (c) \(-(1 / 3)\) (d) \(-3\)

\(\int\left[\left(e^{5 \log x}-e^{3 \log x}\right) /\left(e^{4 \log x}-e^{2 \log x}\right)\right] d x=\) (a) e \(\cdot 2^{-2 \mathrm{x}}\) (b) \(\mathrm{e}^{3} \log _{\mathrm{e}} \mathrm{x}\) (c) \(\left(\mathrm{x}^{3} / 3\right)\) (d) \(\left(\mathrm{x}^{2} / 2\right)\)

\(\int\left[\left\\{\left(x^{5}-x\right)^{1 / 5} d x\right\\} / x^{6}\right]=\ldots+c\) (a) \((5 / 24)\left[1-\left(1 / \mathrm{x}^{4}\right)\right]^{(6 / 5)}\) (b) \((1 / 24)\left[1-\left(1 / \mathrm{x}^{4}\right)\right]^{(1 / 5)}\) (c) \((5 / 24)\left[1-\left(1 / \mathrm{x}^{4}\right)\right]^{(1 / 5)}\) (d) \((5 / 24)\left[1-\left(1 / \mathrm{x}^{4}\right)\right]^{6}\)
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