Inverse substitution method for integrals: integral [from 0 to 9] [sqrt{1 + sqrt{x} }] dx

[weirds111]

Hello, I'm working on a definite integral problem where I have to use inverse substitution to evaluate this definite integral. I have used the method, but for some reason the answer is incorrect compared to the given answer. I would like to know what is wrong with my substitution choice.

I have attached a picture showing my work for this problem. It is right next to the circled number 8.

[MATH]\int_{0}^{9}\sqrt{1+\sqrt{x}}dx [/MATH]
Thanks in advance.

 

[weirds111]

How do I delete this thread? I rechecked my answer and found out the mistake was finding the anti derivative incorrectly. I have no more problems with this problem.

 

[Deleted member 4993]

You should not delete a thread - it might (and will) help another student.

 

[Dr.Peterson]

You made a very simple mistake near the end.

What is the integral of u^4?

It isn't u^4/4.

 

[weirds111]

You made a very simple mistake near the end.

What is the integral of u^4?

It isn't u^4/4.

Yes, I did. Thank you for the reply.

The integral of u^4 is (1/5)u^5

 

[tkhunny]

After writing [math]u > 0[/math], why do you write [math]u = \pm 2\;and\;u = \pm 1 [/math]?

What did you do to show that [math]\dfrac{dx}{du} > 0[/math]?

 

[weirds111]

After writing [math]u > 0[/math], why do you write [math]u = \pm 2\;and\;u = \pm 1 [/math]?

What did you do to show that [math]\dfrac{dx}{du} > 0[/math]?

I understand your point, since I restricted the domain of [math] u [/math] to [math] u > 0 [/math] I should have written down that [math] u = 2\;and\; u = 1 [/math].

I didn't show that [math]\dfrac{dx}{du} > 0[/math] explicitly, but since I restricted the domain of [math] u [/math] to [math] u > 0 [/math] and [math] \dfrac{dx}{du} = 4u^3 - 4u [/math] it is implicit that [math] x [/math] is always increasing. How would I explicitly show that [math]\dfrac{dx}{du} >0[/math]? Should I use the limit notation? For example, [math] lim_{u\to\infty} 4u^3 - 4u = \infty [/math]? therefore [math] x \varepsilon (0, \infty)[/math]

 

[tkhunny]

I understand your point, since I restricted the domain of [math] u [/math] to [math] u > 0 [/math] I should have written down that [math] u = 2\;and\; u = 1 [/math].

I didn't show that [math]\dfrac{dx}{du} > 0[/math] explicitly, but since I restricted the domain of [math] u [/math] to [math] u > 0 [/math] and [math] \dfrac{dx}{du} = 4u^3 - 4u [/math] it is implicit that [math] x [/math] is always increasing. How would I explicitly show that [math]\dfrac{dx}{du} >0[/math]? Should I use the limit notation? For example, [math] lim_{u\to\infty} 4u^3 - 4u = \infty [/math]? therefore [math] x \varepsilon (0, \infty)[/math]

In my mind, it was not immediately obvious that dx/du > 0, since it clearly is not for 0 < u < 1. I just felt a need to know that you thought about it.

 

[weirds111]

In my mind, it was not immediately obvious that dx/du > 0, since it clearly is not for 0 < u < 1. I just felt a need to know that you thought about it.

Oh, I did not notice that dx/du was not increasing from 0 < u < 1. That was a careless mistake for me, thank you for letting me know.