Question

The following integrals can be evaluated only by reversing the order of integration. Sketch the region of integration, reverse the order of integration, and evaluate the integral. $$\int_{0}^{1} \int_{y}^{1} e^{x^{2}} d x d y$$

Short answer

The value of the reversed double integral is given by: $$\frac{\sqrt{\pi}}{2} \text{erf}(1)$$ where erf is the error function.

Step-by-step solution

Sketch the region of integration

The region of integration is given as a planar region defined by the following inequalities: $$0 \leq y \leq 1$$ $$y \leq x \leq 1$$ To draw the region, you can plot these boundaries on the xy-plane and find the enclosed area.

Reverse the order of integration

To reverse the order of integration, first, we need to find the new limits of integration. Since we are going to integrate with respect to y first, we need to find the new boundaries for y. We can find it from the previous inequalities: $$0 \leq x \leq 1$$ $$0 \leq y \leq x$$ Now the integral will become: $$\int_{0}^{1} \int_{0}^{x} e^{x^{2}} d y d x$$

Evaluate the inner integral

Now, we evaluate the inner integral with respect to y: $$\int_{0}^{1} \left[\int_{0}^{x} e^{x^{2}} d y\right] d x$$ Since the integrand doesn't have y, we can consider the e^{x^2} as a constant. $$\int_{0}^{1} \left[ e^{x^{2}} \int_{0}^{x} d y\right] d x$$ Now, we integrate with respect to y: $$\int_{0}^{1} \left[e^{x^{2}} (y) \Big|_{0}^{x}\right] d x$$ $$\int_{0}^{1} e^{x^{2}} x d x$$

Evaluate the outer integral

To find the final solution, we will integrate the remaining integral with respect to x. This integral is a non-elementary integral, meaning, it doesn't have an elementary function as an antiderivative. In this case, we can denote the integral using a special function called the error function (erf). The error function is defined as: $$\text{erf}(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^{2}} dt$$ Recognizing the similarity, rewrite the given integral in terms of the error function, and then we can write the final answer as: $$\frac{\sqrt{\pi}}{2} \text{erf}(x) \Big|_{0}^{1}$$ $$\frac{\sqrt{\pi}}{2} (\text{erf}(1) - \text{erf}(0))$$ $$\frac{\sqrt{\pi}}{2} \text{erf}(1)$$