# Math Help - Reversing the order of integration

1. ## Reversing the order of integration

Question:
Evaluate the iterated integral $\int_0^1{\int_\frac{x}{2}^\frac{1}{2}{e^{y^2}}\,dy }\,dx$ by reversing the order of integration.

Is this correct?
current limits are:
$\frac{x}{2} \leq y \leq \frac{1}{2} = x \leq 2y \leq 1$ and $0 \leq x \leq 1$
rearranging these gives:
$0 \leq y \leq \frac{1}{2}$ and $0 \leq x \leq 2y$

making the equation now $\int_0^\frac{1}{2}{\int_0^{2y}{e^{y^2}}\,dx}\,dy$

Correct? yes/no? Want some reassurance before I try to tackle the integration part!

2. Originally Posted by Dr Zoidburg
Question:
Evaluate the iterated integral $\int_0^1{\int_\frac{x}{2}^\frac{1}{2}{e^{y^2}}\,dy }\,dx$ by reversing the order of integration.

Is this correct?
current limits are:
$\frac{x}{2} \leq y \leq \frac{1}{2} = x \leq 2y \leq 1$ and $0 \leq x \leq 1$
rearranging these gives:
$0 \leq y \leq \frac{1}{2}$ and $0 \leq x \leq 2y$

making the equation now $\int_0^\frac{1}{2}{\int_0^{2y}{e^{y^2}}\,dx}\,dy$

Correct? yes/no? Want some reassurance before I try to tackle the integration part!
This is correct. I always find it a good exercise to draw the domain (whenever possible) so that I can visually see what we're trying to do.

3. cool. Always glad to know I'm on the right track!

For the equation itself, is this correct:
$\int_0^\frac{1}{2}{\int_0^{2y}{e^{y^2}}\,dx}\,dy$
=
$\int_0^\frac{1}{2}{xe^{y^2}\bigg|_{x=0}^{x=2y}}\,d y$
=
$\int_0^\frac{1}{2}{2y*e^{y^2}}\,dy$

setting $u=y^2$ gives $du = 2ydy$
giving:
$\int_{u=0}^{u=\frac{1}{2}}e^u\,du$

$=e^{\frac{1}{2}} - 1 = 0.6487 (4 dp)$

4. Originally Posted by Dr Zoidburg
cool. Always glad to know I'm on the right track!

For the equation itself, is this correct:
$\int_0^\frac{1}{2}{\int_0^{2y}{e^{y^2}}\,dx}\,dy$
=
$\int_0^\frac{1}{2}{xe^{y^2}\bigg|_{x=0}^{x=2y}}\,d y$
=
$\int_0^\frac{1}{2}{2y*e^{y^2}}\,dy$

setting $u=y^2$ gives $du = 2ydy$
giving:
$\int_{u=0}^{u=\frac{1}{2}}e^u\,du$

$=e^{\frac{1}{2}} - 1 = 0.6487 (4 dp)$
Yes, that is correct- although I would have left the answer as $e^{\frac{1}{2}}- 1$ rather than write an approximation.