## hiralpatel121 3 years ago How can I solve the following? (attached)

1. hiralpatel121

2. hiralpatel121

Sample problem:

3. Romero

What step are you confused on the sample problem?

4. hiralpatel121

I understand the sample problem. I just don't know how to use it for the actual problem.

5. Romero

do the same...instead of using 7 use 5

6. hiralpatel121

I did, but it shows up as wrong when I put it in.

7. Romero

then you did something wrong. ...show me the steps

8. Romero

what is the final answer that you got when you did the problem?

9. hiralpatel121

Nevermind, I found my error, instead of 2/9 I put 2/5. Sorry. Can you help me with another problem though?

10. Romero

yes sure....remember to follow all the steps and make sure you don't make a mistake. One small mistake can make the answer wrong

11. hiralpatel121

12. hiralpatel121

Sample:

13. Romero

What's the first step? Let's make the square root into an exponent $\int\limits_{3}^{7}\frac{1}{\sqrt{6x+7}}dx$$\int\limits_{3}^{7}\frac{1}{(6x+7)^\frac{1}{2}}dx$

14. Romero

then we bring the denominator to the top. But in order to do this we have to make the exponent a negative. Then we get $\int\limits_{3}^{7}(6x+7)^{-\frac{1}{2}}dx$

15. Romero

You just integrate what's inside first....can you tell me what you get?

16. hiralpatel121

3x^2+7x+C

17. Romero

No that's not it. Look back in the problem you did before and see how they integrated a square root after it became $(-------)^\frac{1}{2}$

18. Romero

you have to do the chain rule.

19. Romero

Try it one more time....what is the integral?

20. hiralpatel121

I have no idea here..

21. Romero

I thought you know how to integrate something that was a square root. Look at the example problem you did before and see what happens.

22. hiralpatel121

Would it be: 1/12 root of 6x+7?

23. hiralpatel121

Is the answer 1/4? If yes, then I've done my work correct.

24. Romero

I think you made a mistake $\frac{1}{6} *\frac{1}{2}=\frac{1}{12}$ that's how you got the 12 right?

25. hiralpatel121

Yes

26. Romero

it's suppose to be$\frac{1}{6}*2=\frac{2}{3}=\frac{1}{3}$

27. Romero

1/3 root of 6x+7 not 1/12 root of 6x+7

28. hiralpatel121

Oh, so then the answer would be: 1.

29. Romero

what no I got something else

30. Romero

$\frac{1}{3}\sqrt{6x+7}$$\frac{1}{3}\sqrt{6(7)+7}-\frac{1}{3}\sqrt{6(3)+7}$

31. Romero

$\frac{1}{3}(\sqrt{6(7)+7}-\sqrt{6(3)+7})$$\frac{1}{3}(\sqrt{42+7}-\sqrt{18+7})$

32. Romero

can you do the rest?

33. hiralpatel121

Yes, I got it.

34. Romero

what did you get?

35. hiralpatel121

Im doing the math right now.

36. Romero

ok tell me what you get just to check if you got it right.

37. hiralpatel121

2/3

38. Romero

yeah that's what I got good job.

39. Romero

If you have more problems post them in a different question.

40. Romero

I want to give you more medals :) for the good job