Definite Integral, from 0 to 1, of (8x)(sqrt(x^2+4))

[AmySaunders]

Evaluate the integral from 0 to 1 of (8x)(sqrt(x^2+4))



I started out using trigonometric substitution. x=2tan(theta) and dx=2sec^2(theta)d-theta.

I then get the integral from 0 to 1 of (8(2tan(theta)))(2sec^2(theta)) d-theta

I get lost after there. How do I simplify that?

Thank you!

 

[stapel]

Evaluate the integral from 0 to 1 of (8x)(sqrt(x^2+4))

I started out using trigonometric substitution. x=2tan(theta) and dx=2sec^2(theta)d-theta.

Thank you for showing your work and reasoning! But possibly you're overthinking this a bit...?

. . . . .\(\displaystyle \dfrac{d}{dx}\, \sqrt{x^2\, +\, 4\,}\, =\, \left(\dfrac{1}{2}\right) \left(\dfrac{1}{\sqrt{x^2\, +\, 4\,}}\right) \left(2x\right) \, =\, \dfrac{2x}{2\, \sqrt{x^2\, +\, 4\,}}\, =\, \dfrac{x}{\sqrt{x^2\, +\, 4\, }}\)

Can you see where to go with this?

 

[HallsofIvy]

Or, seeing that "\(\displaystyle x^2\,+\, 4\)" in the denominator and "\(\displaystyle x\, dx\)" in the numerator, let \(\displaystyle u\,=\, x^2\,+ \,2\) so that \(\displaystyle du\,= \,2x\,dx\) and \(\displaystyle 8x\,dx\,= \,4\,du\).

\(\displaystyle \displaystyle{\int} \,\dfrac{8x\,dx}{\sqrt{x^2\,+\, 4\,}}\,= \,4\displaystyle{\int}\, \dfrac{du}{\sqrt{u\,}}\,= \,4\,\displaystyle{\int}\, u^{-1/2} \,du\)

 

[AmySaunders]

Thank you for showing your work and reasoning! But possibly you're overthinking this a bit...?

. . . . .\(\displaystyle \dfrac{d}{dx}\, \sqrt{x^2\, +\, 4\,}\, =\, \left(\dfrac{1}{2}\right) \left(\dfrac{1}{\sqrt{x^2\, +\, 4\,}}\right) \left(2x\right) \, =\, \dfrac{2x}{2\, \sqrt{x^2\, +\, 4\,}}\, =\, \dfrac{x}{\sqrt{x^2\, +\, 4\, }}\)

Can you see where to go with this?

So then I can use u substitution and I can set u=x^2+4 which leads to (1/2) times the integral of du/(sqrt(u)) which is (sqrt(u))+C and then I can substitute x back in, getting

(sqrt(x^2+4)) + C as my final answer. What do I do with the (8x)? Can I integrate it separately?

 

[stapel]

. . . . .\(\displaystyle \dfrac{d}{dx}\, \sqrt{x^2\, +\, 4\,}\, =\, \left(\dfrac{1}{2}\right) \left(\dfrac{1}{\sqrt{x^2\, +\, 4\,}}\right) \left(2x\right) \, =\, \dfrac{2x}{2\, \sqrt{x^2\, +\, 4\,}}\, =\, \dfrac{x}{\sqrt{x^2\, +\, 4\, }}\)

Can you see where to go with this?

So then I can use u substitution and I can set u=x^2+4 which leads to (1/2) times the integral of du/(sqrt(u))...

How are you getting this? If \(\displaystyle u\, =\, x^2\, +\, 4,\) then \(\displaystyle du\, =\, 2x\, dx\) so \(\displaystyle 4\, du\, =\, 8x\, dx,\) and:

. . . . .\(\displaystyle \displaystyle{ \int \,} \dfrac{8x}{\sqrt{x^2\, +\, 4\,}} \, dx\, \xrightarrow{sub}\, \displaystyle{ \int \,} \dfrac{4\,du}{\sqrt{u\,}}\)

...which is not the result you're reporting. Instead, why not use the u-substitution indicated in my reply?

. . . . .\(\displaystyle u\, =\, \sqrt{x^2\, +\, 4\,}\)

. . . . .\(\displaystyle du\, =\, \dfrac{x\, dx}{\sqrt{x^2\, +\, 4\, }}\)

The result follows immediately.

 

[Rohit93]

How are you getting this? If \(\displaystyle u\, =\, x^2\, +\, 4,\) then \(\displaystyle du\, =\, 2x\, dx\) so \(\displaystyle 4\, du\, =\, 8x\, dx,\) and:

. . . . .\(\displaystyle \displaystyle{ \int \,} \dfrac{8x}{\sqrt{x^2\, +\, 4\,}} \, dx\, \xrightarrow{sub}\, \displaystyle{ \int \,} \dfrac{4\,du}{\sqrt{u\,}}\)

...which is not the result you're reporting. Instead, why not use the u-substitution indicated in my reply?

. . . . .\(\displaystyle u\, =\, \sqrt{x^2\, +\, 4\,}\)

. . . . .\(\displaystyle du\, =\,\dfrac{x\, dx}{\sqrt{x^2\, +\, 4\, }}\)

The result follows immediately.

can you please elaborate on how we get ???
\(\displaystyle du\, =\, \frac{x\, dx}{\sqrt{x^2+4}}\)
after differentiating .\(\displaystyle u\, =\, \sqrt{x^2\, +\, 4\,}\)

 

[pka]

Evaluate the integral from 0 to 1 of (8x)(sqrt(x^2+4)

I simply do not understand what any of you are doing. As as far as I can see, no substitution is needed.

In the OP the question is about \(\displaystyle \displaystyle\int_0^1 {\left( {8x} \right)\sqrt {{x^2} + 4} dx} \)

Where is there denominator?

QUESTION: What is the derivative of \(\displaystyle \displaystyle~\frac{8}{3}\sqrt {{{\left( {{x^2} + 4} \right)}^3}} ~?\)

 

[stapel]

. . . . .\(\displaystyle \dfrac{d}{dx}\, \sqrt{x^2\, +\, 4\,}\, =\, \left(\dfrac{1}{2}\right) \left(\dfrac{1}{\sqrt{x^2\, +\, 4\,}}\right) \left(2x\right) \, =\, \dfrac{2x}{2\, \sqrt{x^2\, +\, 4\,}}\, =\, \dfrac{x}{\sqrt{x^2\, +\, 4\, }}\)

Can you see where to go with this?

...why not use the u-substitution indicated in my reply?

. . . . .\(\displaystyle u\, =\, \sqrt{x^2\, +\, 4\,}\)

. . . . .\(\displaystyle du\, =\,\dfrac{x\, dx}{\sqrt{x^2\, +\, 4\, }}\)

can you please elaborate on how we get ???
\(\displaystyle du\, =\, \frac{x\, dx}{\sqrt{x^2+4}}\)
after differentiating .\(\displaystyle u\, =\, \sqrt{x^2\, +\, 4\,}\)

Which part of the step-by-step work I showed is causing you confusion? What divergent result did you obtain when you took the derivative yourself?

Please be specific and complete, showing all of your steps. Thank you!

 

[Rohit93]

Which part of the step-by-step work I showed is causing you confusion? What divergent result did you obtain when you took the derivative yourself?

Please be specific and complete, showing all of your steps. Thank you!

hi stapel,
i have the difficulty of understanding how the term \(\displaystyle \sqrt{x^2+4}\) end up in the denominator

 

[stapel]

. . . . .\(\displaystyle \dfrac{d}{dx}\, \sqrt{x^2\, +\, 4\,}\, =\, \left(\dfrac{1}{2}\right) \left(\dfrac{1}{\sqrt{x^2\, +\, 4\,}}\right) \left(2x\right) \, =\, \dfrac{2x}{2\, \sqrt{x^2\, +\, 4\,}}\, =\, \dfrac{x}{\sqrt{x^2\, +\, 4\, }}\)

hi stapel,
i have the difficulty of understanding how the term \(\displaystyle \sqrt{x^2+4}\) end up in the denominator

Anything inside a square root may also be regarded as being "to the one-half power". What did you get when you did the differentiation of the square root?

Please show all of your steps, just as I did (as quoted above), so we can see where things are going sideways. Thank you!