## roselin Group Title need help with :definite integral from 0 to 1 x^3/ Square root of x^4+9 one year ago one year ago

1. ChmE Group Title

$\int\limits_{0}^{1}{x^3 \over \sqrt{x^4+9}}dx$Is this the correct equation?

2. roselin Group Title

yes, that is the one.

3. ChmE Group Title

u-substitute for x^4 + 9

4. roselin Group Title

let u be x^4+9?

5. roselin Group Title

but how do you know which one to put as" u"?

6. ChmE Group Title

u = x^4 + 9 du = 4x^3 dx du / 4x^3 = dx$\int\limits\limits_{0}^{1}{x^3 \over \sqrt{u}}{du \over 4x^3}$$\int\limits\limits_{0}^{1}{du \over 4\sqrt{u}}$Can you finish it? I think you are actually supposed to re-evaluate the limits but I don't remember how to do that. It is not necessary if you replace u with what you substituted it for before you apply the limits

7. roselin Group Title

let me try replacing it

8. roselin Group Title

|dw:1355885411954:dw|

9. TweT226 Group Title

to re-evaluate the limits plug in 0 and 1 into u=x^4 + 9

10. roselin Group Title

then, what do I need to do after that?

11. ChmE Group Title

I don't know why you did that.. you need to integrate the last equation I gave you that has u in it

12. ChmE Group Title

$\int\limits\limits\limits_{9}^{10}{du \over 4\sqrt{u}}$

13. ChmE Group Title

@TweT226 thx

14. ChmE Group Title

After you do u-substitution to re-evaluate the limits see Twe's post

15. roselin Group Title

when i plug in 0 i get 9 and whne i plug in 1 , I get 10

16. ChmE Group Title

$\int\limits\limits\limits\limits_{9}^{10}{1 \over 4}u^{-1/2}du$

17. roselin Group Title

how did u get u^-1/2?

18. ChmE Group Title

${{1 \over 4}u^{1/2} \over 1/2}|_9^{10}$${1 \over 2}u^{1/2}|_9^{10}$${1 \over 2}10^{1/2}-{1 \over 2}9^{1/2}$

19. ChmE Group Title

$\sqrt x=x^{1/2}$${1 \over x^n}=x^{-n}$

20. roselin Group Title

oh,okay.i see

21. ChmE Group Title

re-evaluating the limits is the short method. If you don't do that you can plug what you substituted u for then use the old limits after integration. You should get the same answer

22. roselin Group Title

okay.

23. ChmE Group Title

$\int\limits\limits_{0}^{1}{x^3 \over \sqrt{x^4+9}}dx$u-substitution: u = x^4 + 9 du = 4x^3 dx du / 4x^3 = dx Re-evaluating the limits u = (0)^4 + 9 = 9 u = (1)^4 + 9 = 10$\int\limits\limits\limits_{9}^{10}{x^3 \over \sqrt{u}}{du \over 4x^3}$$\int\limits\limits\limits\limits_{9}^{10}{du \over 4\sqrt{u}}$$\int\limits\limits\limits\limits\limits_{9}^{10}{1 \over 4}u^{-1/2}du$${{1 \over 4}u^{1/2} \over 1/2}|_9^{10}$${1 \over 2}u^{1/2}|_9^{10}$${1 \over 2}10^{1/2}-{1 \over 2}9^{1/2}$

24. roselin Group Title

thanks