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Definite integral via u-substitution

Evaluate the definite integral using uu-substitution.

Evaluate the definite integral

∫02xe−x2 dx\int_0^2 x e^{-x^2}\,dx

using the substitution u=−x2u = -x^2.

Hint

Let u=−x2u = -x^2. Find dudu in terms of x dxx\,dx, then rewrite the bounds in terms of uu before integrating.

Step-by-step solution

Let u=−x2u = -x^2, so du=−2x dxdu = -2x\,dx, which gives x dx=−12dux\,dx = -\tfrac{1}{2}du. When x=0x=0, u=0u=0, and when x=2x=2, u=−4u=-4. Substituting transforms the integral into ∫0−4eu(−12)du\int_0^{-4} e^u \left(-\tfrac{1}{2}\right) du.

Pulling out the constant gives −12∫0−4eu du=−12[eu]0−4=−12(e−4−e0)-\tfrac{1}{2}\int_0^{-4} e^u\,du = -\tfrac{1}{2}\left[e^u\right]_0^{-4} = -\tfrac{1}{2}\left(e^{-4} - e^0\right). Simplifying, this equals 12(1−e−4)\tfrac{1}{2}\left(1 - e^{-4}\right).

Direction of bound transformation in u-substitution

Hint
Step-by-step solution

The new upper bound was less than the new lower bound. When x equals zero, u equals zero, and when x equals two, u equals negative four. Since negative four is less than zero, the transformed integral runs from zero down to negative four, reversing the usual order.

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