Definite Integration
Grade 12
Question:
<p>Evaluate \(\displaystyle\iint_{[0,1]^2}e^{\max(x^2,y^2)}\,dx\,dy\) [JEE Advanced 2008]</p>
\(\dfrac{e-1}{2}+\dfrac{e}{2}\)
\(e-\dfrac{1}{2}\)
\(\dfrac{e}{2}\)
\(e-1\)
Step-by-Step Solution
Key Concept: By symmetry: integral = 2\int_0^1\int_0^x e^(x^2)dy dx + ... split region where x>y and x<y.
<div class='solution'>
<p>By symmetry: $\iint = 2\int_0^1\int_0^x e^{x^2}\,dy\,dx = 2\int_0^1 xe^{x^2}\,dx = 2\cdot\frac{e-1}{2}=e-1$.</p>
<p>Wait: we need to include the diagonal line x=y (measure zero). The two regions: $\{x>y\}$ and $\{x<y\}$. On $\{x>y\}$: max=x^2. On $\{x<y\}$: max=y^2. By symmetry equal.</p>
<p>$I=2\int_0^1\int_0^x e^{x^2}dy\,dx=2\int_0^1 xe^{x^2}dx=[e^{x^2}]_0^1=e-1$.</p>
<p>Hmm but option D=e-1 and option B=e-1/2. The correct computation gives e-1. Checking again: $2\int_0^1 xe^{x^2}dx = 2\cdot\frac{e-1}{2}=e-1$. Accept D=e-1.</p>
Correct Answer: B