Question

evaluate the integral.
int_{1}^{3} x e^{-x^{2}} d x
int_{1}^{3} x e^{-x^{2}} d x=quad
(type an integer or decimal rounded to three decimal places as needed.)

Explanation:

Step1: Use substitution

Let \(u = -x^{2}\), then \(\mathrm{d}u=-2x\mathrm{d}x\), and \(x\mathrm{d}x=-\frac{1}{2}\mathrm{d}u\).
When \(x = 1\), \(u=-1\); when \(x = 3\), \(u=-9\).
The integral \(\int_{1}^{3}xe^{-x^{2}}\mathrm{d}x\) becomes \(-\frac{1}{2}\int_{-1}^{-9}e^{u}\mathrm{d}u\).

Step2: Integrate \(e^{u}\)

The integral of \(e^{u}\) with respect to \(u\) is \(e^{u}\). So \(-\frac{1}{2}\int_{-1}^{-9}e^{u}\mathrm{d}u=-\frac{1}{2}[e^{u}]_{-1}^{-9}\).

Step3: Evaluate the definite - integral

Using the fundamental theorem of calculus \(F(b)-F(a)\), we have \(-\frac{1}{2}(e^{-9}-e^{-1})\).

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Since \(e\approx2.718\), \(e^{9}\approx8103.084\)

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Answer:

\(0.184\)