Dido525 2 years ago Evaluate the intergral:

1. Dido525

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2. Dido525

I thought I could use substitution but I can't ...

3. winterfez

is this right? $\int\limits_{0}^{4}\frac{ x }{ \sqrt{1+2x} }$

4. Dido525

yep :) .

5. eliassaab

put u= 1 + 2 x

6. Mimi_x3

$$u = 1+ 2x => du/dx = 2$$ $$1+2x = u => 2x = u-1 => x = (u-1)/2$$

7. Dido525

I got to that step. Then I got stuck :P .

8. Mimi_x3

Maybe:$\int\limits\frac{x}{\sqrt{1+2x}} dx => \int\limits \frac{(u-1)/2}{\sqrt{u}} *2du \int\limits\frac{u-1}{2} *\frac{1}{\sqrt{u}} 2du => 2\int\limits\frac{u-1}{2} *\frac{1}{\sqrt{u}}$

9. eliassaab

Here what wolfram alpha gave

10. Mimi_x3

Would my step work tho?

11. zepdrix

@Dido525 do you understand what mimi did? those are the correct steps to take.

12. Dido525

Yeah I actually do. I just need to process it.

13. Dido525

@Mimi_x3 : Wouldn't you write du/2 instead of 2*du ?

14. Dido525

Because we said du/2 was dx.

15. zepdrix

Oh yes :o good call.

16. Mimi_x3

Yeah typo lol sorry! that was where i went wrong :(

17. Dido525

It's okay :) .

18. Dido525

So so far I have: $\frac{ 1 }{ 4 }\int\limits_{1}^{9} \frac{ u-1 }{ \sqrt{u} } du$

19. Mimi_x3

then break it up!

20. Dido525

Ohh... Right :P .

21. Mimi_x3

integrate it; then don't forget to sub back in $$u$$

22. Dido525

Thanks! :) .

23. Mimi_x3

:)