Calculate the following integral: integral from 0
Calculate the following integral: $$ \int_0^a \int_x^a e^{y^2}\,dy\,dx. $$
Explanation of Method
$e^{y^2}$ has no elementary antiderivative, so integrating with respect to $y$ first is a dead end. Reversing the order makes $y$ the outer variable and the inner integral (with respect to $x$) trivial, since $e^{y^2}$ is then just a constant with respect to $x$.
- Identify the triangular region and reverse the order.
- Integrate with respect to $x$ (trivial, since $e^{y^2}$ doesn't depend on $x$).
- Integrate $ye^{y^2}$ with respect to $y$ via substitution $u=y^2$.
Step 1 · Identify the region
From $\int_0^a\int_x^a e^{y^2}\,dy\,dx$: $0\leqslant x\leqslant a$, $x\leqslant y\leqslant a$. This is the triangle with vertices $(0,0)$, $(0,a)$, $(a,a)$: bounded left by $x=0$, right/above by $y=a$, and below by $y=x$.
Step 2 · Reverse the order of integration
With $y$ as the outer variable: $0\leqslant y\leqslant a$, $0\leqslant x\leqslant y$.
Step 3 · Rewrite the integral
$$ I = \int_0^a\int_0^y e^{y^2}\,dx\,dy. $$
Step 4 · Integrate with respect to $x$
$$ \int_0^y e^{y^2}\,dx = e^{y^2}\cdot[x]_0^y = ye^{y^2}. $$
Step 5 · Integrate with respect to $y$
$$ I = \int_0^a ye^{y^2}\,dy. $$
Let $u=y^2$, $du=2y\,dy$; when $y=0$, $u=0$; when $y=a$, $u=a^2$:
$$ I = \frac12\int_0^{a^2} e^u\,du = \frac12\left[e^u\right]_0^{a^2} = \frac{e^{a^2}-1}{2}. $$
Additional Notes
- Whenever the inner integral involves a function of the outer-integration variable with no elementary antiderivative (like $e^{y^2}$, or $\sin(x^2)$), try reversing the order — it may become the constant factor of a trivial inner integral instead.
Result
The solution to the integral is: