double integral of x(1+y^2)^-0.5 over region in first quadrant bounded by

bounded by y=x^2, y=4, x=0. picture: graph y = x^2, y=4, and x=0 from x=0 to x=2 - Wolfram|Alpha

i set my integral: integrate 0 to 2 [ integrate x^2 to 4 x(1+y^2)^-0.5 dy ] dx

then I tried doing it, first integral results: x[arcsinh(y)] from 4 to x^2

but second integral, how to integrate x*arcsinh(x^2).

edit: is this how

[[x^2⋅arcsinh(x^2)] - [(x^4 +1)^.5]]/2

Re: double integral of x(1+y^2)^-0.5 over region in first quadrant bounded by

Quote:

Originally Posted by

**SNAKE** bounded by y=x^2, y=4, x=0. picture:

graph y = x^2, y=4, and x=0 from x=0 to x=2 - Wolfram|Alpha
i set my integral: integrate 0 to 2 [ integrate x^2 to 4 x(1+y^2)^-0.5 dy ] dx

then I tried doing it, first integral results: x[arcsinh(y)] from 4 to x^2

but second integral, how to integrate x*arcsinh(x^2).

edit: is this how

[[x^2⋅arcsinh(x^2)] - [(x^4 +1)^.5]]/2

reverse the order of integration?

Re: double integral of x(1+y^2)^-0.5 over region in first quadrant bounded by

Quote:

Originally Posted by

**SNAKE** bounded by y=x^2, y=4, x=0. picture:

graph y = x^2, y=4, and x=0 from x=0 to x=2 - Wolfram|Alpha
i set my integral: integrate 0 to 2 [ integrate x^2 to 4 x(1+y^2)^-0.5 dy ] dx

then I tried doing it, first integral results: x[arcsinh(y)] from 4 to x^2

but second integral, how to integrate x*arcsinh(x^2).

edit: is this how

[[x^2⋅arcsinh(x^2)] - [(x^4 +1)^.5]]/2

Horizontal strips are a better choice than vertical in this case. So $\displaystyle \begin{align*} 0 \leq x \leq \sqrt{y} \end{align*}$ and $\displaystyle \begin{align*} 0 \leq y \leq 4 \end{align*}$, so that means your double integral is

$\displaystyle \begin{align*} \int_0^4{\int_0^{\sqrt{y}}{x \left( 1 + y^2 \right) ^{-\frac{1}{2}} \,dx}\,dy} &= \int_0^4{ \left[ \frac{x^2}{2} \left( 1 + y^2 \right) ^{-\frac{1}{2}} \right]_0^{\sqrt{y}} \,dy } \\ &= \int_0^4{ \left[ \frac{ \left( \sqrt{y} \right) ^2 }{2} \left( 1 + y^2 \right) ^{-\frac{1}{2}} \right] - \left[ \frac{0^2}{2} \left( 1 + y^2 \right) ^{-\frac{1}{2}} \right] \,dy } \\ &= \frac{1}{2}\int_0^4{ y \, \left( 1 + y^2 \right) ^{-\frac{1}{2}} \, dy }\\ &= \frac{1}{4} \int_0^4{ 2y \, \left( 1 + y^2 \right) ^{-\frac{1}{2}} \, dy } \\ &= \frac{1}{4} \int_1^{17}{ u^{-\frac{1}{2}} \, du } \textrm{ when we substitute } u =1 + y^2 \\ &= \frac{1}{4} \left[ 2u^{\frac{1}{2}} \right]_1^{17} \\ &= \frac{1}{4} \left[ 2 \left( 17 \right) ^{\frac{1}{2}} - 2 \left( 1 \right) ^{\frac{1}{2}} \right] \\ &= \frac{1}{2} \left( \sqrt{17} - 1 \right) \end{align*}$

Re: double integral of x(1+y^2)^-0.5 over region in first quadrant bounded by

ahh, i considered doing the horizontal method but I didn't see how it would alleviate the difficult integral but seeing it written out like this, you have Proven It!!

I am adding extra smilies to my sig to show my appreciation :)

Re: double integral of x(1+y^2)^-0.5 over region in first quadrant bounded by

Quote:

Originally Posted by

**SNAKE** bounded by y=x^2, y=4, x=0. picture:

graph y = x^2, y=4, and x=0 from x=0 to x=2 - Wolfram|Alpha
i set my integral: integrate 0 to 2 [ integrate x^2 to 4 x(1+y^2)^-0.5 dy ] dx

then I tried doing it, first integral results: x[arcsinh(y)] from 4 to x^2

but second integral, how to integrate x*arcsinh(x^2).

edit: is this how

[[x^2⋅arcsinh(x^2)] - [(x^4 +1)^.5]]/2

Reversing the order of integration, as both Idea and Prove It said is the best way to handle this integral. But to answer you last question, "how to integrate x arcsinh(x^2)" let u= x^2 so [tex]\int x arcsinh(x) dx[tex] becomes $\displaystyle \frac{1}{2}\int arcsinh(u)du$ and that can be looked up in a table of integrals.

(Lists of integrals - Wikipedia, the free encyclopedia)

Re: double integral of x(1+y^2)^-0.5 over region in first quadrant bounded by

Quote:

Originally Posted by

**HallsofIvy** Reversing the order of integration, as both Idea and Prove It said is the best way to handle this integral. But to answer you last question, "how to integrate x arcsinh(x^2)" let u= x^2 so [tex]\int x arcsinh(x) dx[tex] becomes $\displaystyle \frac{1}{2}\int arcsinh(u)du$ and that can be looked up in a table of integrals.

(

Lists of integrals - Wikipedia, the free encyclopedia)

Or otherwise derived (note I've left out the integration constants until the end)...

Let's consider $\displaystyle \begin{align*} \int{ \frac{2x}{\sqrt{1 + x^2}} \, dx} = \int{ 2x \left( 1 + x^2 \right) ^{-\frac{1}{2}} \,dx} \end{align*}$. With the substitution $\displaystyle \begin{align*} u = 1 + x^2 \implies du = 2x\,dx \end{align*}$ we can see $\displaystyle \begin{align*} \int{ 2x \left( 1 + x^2 \right) ^{-\frac{1}{2}} \, dx} = \int{ u^{-\frac{1}{2}} \, du } = 2u^{\frac{1}{2}} = 2 \, \sqrt{ 1 + x^2 } \end{align*}$.

We might also consider integration by parts:

$\displaystyle \begin{align*} \int{ \frac{2x}{\sqrt{1 + x^2}} \, dx} &= \int{ 2x \, \frac{1}{\sqrt{1 + x^2}} \, dx} \\ &= 2x\,\textrm{arsinh}\,{(x)} - \int{ 2\,\textrm{arsinh}\,{(x)}\,dx } \\ &= 2x\,\textrm{arsinh}\,{(x)} - 2\int{ \textrm{arsinh}\,{(x)}\,dx } \end{align*}$

and thus

$\displaystyle \begin{align*} 2x\,\textrm{arsinh}\,{(x)} - 2\int{ \textrm{arsinh}\,{(x)}\,dx} &= 2\,\sqrt{1 + x^2} \\ x\,\textrm{arsinh}\,{(x)} - \int{ \textrm{arsinh}\,{(x)}\,dx} &= \sqrt{1 + x^2} \\ \int{\textrm{arsinh}\,{(x)}\,dx} &= x\,\textrm{arsinh}\,{(x)} - \sqrt{1 + x^2} + C \end{align*}$