MHB Problem of the Week #98 - February 10th, 2014

[Chris L T521]

Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: Evaluate the integral
\[\int_0^1 \int_0^1 \exp(\max\{x^2,y^2\})\,dy\,dx\]
where $\exp(u) = e^u$ and $\max\{x^2,y^2\}$ means the larger of the numbers $x^2$ and $y^2$.

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[Chris L T521]

This week's problem was correctly answered by chisigma, lfdahl, magneto, MarkFL, and Pranav. You can find Mark's solution below.

[sp]First, we need to determine where $y^2>x^2$ in the square region over which we are integrating. We find that the diagonal of the square along the line $y=x$ divides the region into two areas. The right isosceles triangle with vertices $(0,0),\,(0,1),\,(1,1)$ is where $y^2\ge x$ and the right isosceles triangle with vertices $(0,0),\,(1,1),\,(1,0)$ is where $x^2\ge y^2$.

We will treat the triangular region where $x^2\ge y^2$ as a Type I region and the region where $y^2\ge x$ as a Type II region.

Thus, we may state:

$$I=\int_0^1\int_0^1 \exp\left(\max\left(x^2,y^2 \right) \right)\,dy\,dx=\int_0^1\int_0^x e^{x^2}\,dy\,dx+\int_0^1\int_0^y e^{y^2}\,dx\,dy$$

Rewriting the iterated integrals, we have:

$$I=\int_0^1 e^{x^2}\left[\int_0^x \,dy \right]\,dx+\int_0^1e^{y^2}\left[\int_0^y \,dx \right]\,dy$$

Applying the FTOC on the inner integrals, we obtain:

$$I=\int_0^1 xe^{x^2}\,dx+\int_0^1 ye^{y^2}\,dy$$

Observing that if we change the dummy variables of integration both to $u$, we find:

$$I=2\int_0^1 ue^{u^2}\,du$$

Using the substitution:

$$v=u^2\,\therefore\,dv=2u\,du$$

We find:

$$I=\int_0^1 e^{v}\,dv=\left.e^v \right|_0^1=e-1$$

Hence, we may conclude that:

$$I=\int_0^1\int_0^1 \exp\left(\max\left(x^2,y^2 \right) \right)\,dy\,dx=e-1$$[/sp]