Question

Evaluate \(\int_{2}^{3} \exp \left[-2(x-3)^{2}\right] d x\)

Short answer

The evaluated integral is \(-\frac{\sqrt{\pi}}{4}(erf(0) - erf(\sqrt{2}))\).

Step-by-step solution

Identify a suitable substitution

Firstly, observe that the exponent of the function exponent is \([-2(x-3)^{2}]\). To simplify this, a good substitution to make would be \(u = (x-3)\). Therefore, \(du=dx\).

Apply the substitution

Substitute \(u\) and \(du\) into the integral and adjust the limits of the integral. The limits change from 2 to 3 to -1 to 0. The integral becomes \(\int_{-1}^{0} e^{-2u^{2}} du\).

Solve the integral

This integral can not be solved using elementary functions. However, it can be expressed in terms of a special function known as the error function, often abbreviated as erf. Thus, the integral is equal to \(-\frac{\sqrt{\pi}}{4}[erf(-\sqrt{2}u)]_{-1}^{0}\).

Substitute the limits

Substitute the limits into the error function and simplify. We get, \(-\frac{\sqrt{\pi}}{4}(erf(0) - erf(\sqrt{2})).\