Question

evaluate the integral.
int_{1}^{2} x e^{-x^{2}} d x
int_{1}^{2} 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 \(du=-2x dx\), and \(x dx=-\frac{1}{2}du\).
When \(x = 1\), \(u=-1\); when \(x = 2\), \(u=-4\).
The integral \(\int_{1}^{2}xe^{-x^{2}}dx\) becomes \(-\frac{1}{2}\int_{-1}^{-4}e^{u}du\).

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

The integral of \(e^{u}\) with respect to \(u\) is \(e^{u}\). So \(-\frac{1}{2}\int_{-1}^{-4}e^{u}du=-\frac{1}{2}[e^{u}]_{-1}^{-4}\).

Step3: Evaluate the definite - integral

Using the fundamental theorem of calculus \(F(b)-F(a)\), we have \(-\frac{1}{2}(e^{-4}-e^{-1})\).

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Since \(e\approx2.718\), \(e^{4}\approx54.598\)

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Answer:

\(0.175\)