15.2.60 Reverse the order of integration

[karush]

$\textsf{Reverse the order of integration in the following integral }$

\begin{align*}\displaystyle
I&=\int_0^1 \int_2^{2e^x}f(x,y) dydx
\end{align*}
ok tried to follow some examples but
\begin{align*}\displaystyle
I&=\int_0^1 \int_?^{?}f(x,y) dxdy
\end{align*}

 

[mathmari]

From the integral we have that $0\leq x\leq 1$ and $2\leq y\leq 2e^x$.

So, we get that $y\leq 2e^x \Rightarrow \frac{y}{2}\leq e^x \Rightarrow \ln \left (\frac{y}{2}\right )\leq x$.

Therefore, we get that $\ln \left (\frac{y}{2}\right )\leq x\leq 1$ and $2\leq y\leq 2e^x\leq 2e^1=2e$.

We have to check if the condition $0 \le x$ still holds. We have that $\ln \left( \frac y 2\right) \ge 0$, so the condition holds.

So, by changing the order of integrals we get the following $$I=\int_2^{2e} \int_{\ln \left (\frac{y}{2}\right )}^1f(x,y) dxdy $$

 

[karush]

ok let me try one more that is similar
Reverse the order of integration in the following integral
$\displaystyle\int_{0}^{6} \int_{\sin^{-1}{\frac{y}{6}}}^{^{\frac{\pi}{2}}} f(x,y) \, dx \,dy$
$\textsf{From the integral we have that $0 \le x \le 6$}$
and
$\sin^{-1}{\frac{y}{6}} \le y \le \frac{\pi}{2}$
$\therefore y \le \frac{\pi}{2}$ and $\sin^{-1}{\frac{y}{6}} \le y$

do we plug $6$ into $\displaystyle \sin^{-1}{\frac{y}{6}}\implies \sin^{-1}{\frac{6}{6}}\implies \frac{\pi}{2}$ ?

 

[mathmari]

From $\displaystyle\int_{0}^{6} \int_{\sin^{-1}{\frac{y}{6}}}^{\frac{\pi}{2}} f(x,y) \, dx \,dy$ we have that the inner integral corresponds to the inner differential, $dx$, and the outer intergral to $dy$.

If we use parentheses it will be clearer: $$\int_{0}^{6} \left (\int_{\sin^{-1}{\frac{y}{6}}}^{\frac{\pi}{2}} f(x,y) \, dx \right )\,dy$$ Therefore, we have that $0\leq y\leq 6$ and $\sin^{-1}\frac{y}{6}\leq x\leq \frac{\pi}{2}$.

Try from here to find the new limits.

 

[karush]

$$\int_{0}^{6} \left (\int_{\sin^{-1}{\frac{y}{6}}}^{\frac{\pi}{2}} f(x,y) \, dx \right )\,dy$$
Therefore, we have that $0\leq y\leq 6$ and $\sin^{-1}\frac{y}{6}\leq x\leq \frac{\pi}{2}$.

$\sin^{-1}\frac{y}{6}= x
\implies\sin(x)=\frac{y}{6}
\implies 6\sin(x)=y $
so does $\frac{\pi}{2}=x$.

 

[mathmari]

$\sin^{-1}\frac{y}{6}= x
\implies\sin(x)=\frac{y}{6}
\implies 6\sin(x)=y $
so does $\frac{\pi}{2}=x$.

We have that $0\leq y\leq 6$ and $\sin^{-1}\frac{y}{6}\leq x\leq \frac{\pi}{2}$

We get $\sin^{-1}\frac{y}{6}\leq x \Rightarrow \frac{y}{6}\leq \sin (x) \Rightarrow y\leq 6\sin (x)$ and so $0\leq y\leq 6\sin (x)$ and $0\leq x\leq \frac{\pi}{2}$.

So, we get the double integral $\int_{0}^{\frac{\pi}{2}} \left (\int_{0}^{6\sin (x)} f(x,y) \, dy \right )\,dx$.