# Thread: laws for changing integrating orger

1. ## laws for changing integrating orger

$\displaystyle \int_{0}^{2}\int_{y^{2}}^{\sqrt[3]{y^{2}}}ye^{x^{2}}dxdy$

we solve it by changing the order of ontegration
at this integral
y goes from 0 till 2
x goes from $\displaystyle x=y^{2}$ till $\displaystyle x=\sqrt[3]{y^{2}}$
now we what first integrate by y then by x
so when we integrate by y: our intervals are $\displaystyle y=\sqrt{x}$ till $\displaystyle y=\sqrt[2]{y^{3}}$

what are the intervals when we integrate by x

?

2. ## Re: laws for changing integrating orger

Originally Posted by transgalactic
$\displaystyle \int_{0}^{2}\int_{y^{2}}^{\sqrt[3]{y^{2}}}ye^{x^{2}}dxdy$

we solve it by changing the order of ontegration
at this integral
y goes from 0 till 2
x goes from $\displaystyle x=y^{2}$ till $\displaystyle x=\sqrt[3]{y^{2}}$
now we what first integrate by y then by x
so when we integrate by y: our intervals are $\displaystyle y=\sqrt{x}$ till $\displaystyle y=\sqrt[2]{y^{3}}$

what are the intervals when we integrate by x

?
Did you draw the region defined by the terminals of the given double integral? It should then be obvious.

3. ## Re: laws for changing integrating orger

if i draw it than it will be obvios

but not every formula i could draw and i dont want to remmember every type

how to solve it mathematickly without using graph

4. ## Re: laws for changing integrating orger

Originally Posted by transgalactic
$\displaystyle \int_{0}^{2}\int_{y^{2}}^{\sqrt[3]{y^{2}}}ye^{x^{2}}dxdy$

we solve it by changing the order of ontegration
at this integral
y goes from 0 till 2
x goes from $\displaystyle x=y^{2}$ till $\displaystyle x=\sqrt[3]{y^{2}}$
now we what first integrate by y then by x
so when we integrate by y: our intervals are $\displaystyle y=\sqrt{x}$ till $\displaystyle y=\sqrt[2]{y^{3}}$

what are the intervals when we integrate by x

?
Your terminals tell you $\displaystyle \displaystyle y^2 \leq x \leq \sqrt[3]{y^2}$ and $\displaystyle \displaystyle 0 \leq y \leq 2$.

From the second inequality

$\displaystyle \displaystyle 0 \leq y \leq 2 \implies 0 \leq y^2 \implies 0 \leq x$

and

$\displaystyle \displaystyle 0 \leq y \leq 2 \implies \sqrt[3]{y^2} \leq \sqrt[3]{4} \implies x \leq \sqrt[3]{4}$.

So you have new $\displaystyle \displaystyle x$ terminals $\displaystyle \displaystyle 0 \leq x \leq \sqrt[3]{4}$.

Now from the first inequality

$\displaystyle \displaystyle y^2 \leq x \leq \sqrt[3]{y^2} \implies y \leq \sqrt{x}$

and

$\displaystyle \displaystyle y^2 \leq x \leq \sqrt[3]{y^2} \implies \sqrt{x^3} \leq y$.

So you now have $\displaystyle \displaystyle y$ terminals $\displaystyle \displaystyle \sqrt{x^3} \leq y \leq \sqrt{x}$.

Therefore

$\displaystyle \displaystyle \int_0^2{\int_{y^2}^{\sqrt[3]{y^2}}{y\,e^{x^2}\,dx}\,dy} = \int_0^{\sqrt[3]{4}}{\int_{\sqrt{x^3}}^{\sqrt{x}}{y\,e^{x^2}\,dy}\ ,dx}$

cool proof
thank