# Double integral $\int{e^{t^2}}$over a triangle

Evaluate the following integral $$\int\int_D e^{{t}^{2}} dt~du$$ where $D$ is the triangle lying between the lines $u=0$, $u=t$, $t=1$

Now I understand that $\int e^{t^2}dt$ can never be solved analytically, so I'm guessing this question is going to involve some sort of change of limits?

From the given triangle, I have worked out that (tell me if this is right lol): $$D = \{(u,t) : 0\leq u\leq 1,~0\leq t \leq u \}$$ and that the triangle has coordinates $(1,0), (0,0) \text{ and } (1,1)$

So I would imagine that now the integral becomes: $$\int_0^1 \int_0^u e^{t^2} dt~du~~?$$

But I still can't figure out how to do the integral. So am I getting somewhere?If I'm then how am suppose to carry on?

• I think you've got the wrong triangle. You most likely want $$D=\{(u,t):0\le u\le t, 0\le t\le 1\}$$ instead. Commented Mar 21, 2017 at 0:17

$\int\limits_{t=0}^1 \int\limits_{u=0}^t e^{t^2}\ dt\ du = {e - 1 \over 2}$.
• Can you elaborate on how you did the integral $\int_0^t e^{t^2} dt$ Commented Mar 21, 2017 at 8:35
In the D, you can do the integration with respect to $t$ first or $u$ first, and the result is identical. For the first case, select an arbitrary fixed $u$ and draw a horizontal line across the D and find the intersects, namely $u$ and $1$, and the two values of t are the limits of inner integral. And where is u? Just sweeping the $u$ through the entire region of D, you can find $u$ goes from $0$ to 1, which is the limits for outer integral. For the second case, just refers to the case one. $$\int_0^1(\int_u^1e^{t^2}dt)du=\int_0^1(\int_0^te^{t^2}du)dt=\frac{\sqrt{\pi}}{2} Erfi(1)=1.46$$ $Erfi(x)=\int_0^te^{t^2}dt$ is the imaginary error function.