## hiralpatel121 Group Title How can I solve the following? (attached) 2 years ago 2 years ago

1. hiralpatel121 Group Title

2. hiralpatel121 Group Title

Sample problem:

3. Romero Group Title

What step are you confused on the sample problem?

4. hiralpatel121 Group Title

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

5. Romero Group Title

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

6. hiralpatel121 Group Title

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

7. Romero Group Title

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

8. Romero Group Title

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

9. hiralpatel121 Group Title

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

10. Romero Group Title

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 Group Title

12. hiralpatel121 Group Title

Sample:

13. Romero Group Title

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 Group Title

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 Group Title

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

16. hiralpatel121 Group Title

3x^2+7x+C

17. Romero Group Title

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 Group Title

you have to do the chain rule.

19. Romero Group Title

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

20. hiralpatel121 Group Title

I have no idea here..

21. Romero Group Title

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 Group Title

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

23. hiralpatel121 Group Title

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

24. Romero Group Title

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 Group Title

Yes

26. Romero Group Title

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

27. Romero Group Title

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

28. hiralpatel121 Group Title

Oh, so then the answer would be: 1.

29. Romero Group Title

what no I got something else

30. Romero Group Title

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

31. Romero Group Title

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

32. Romero Group Title

can you do the rest?

33. hiralpatel121 Group Title

Yes, I got it.

34. Romero Group Title

what did you get?

35. hiralpatel121 Group Title

Im doing the math right now.

36. Romero Group Title

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

37. hiralpatel121 Group Title

2/3

38. Romero Group Title

yeah that's what I got good job.

39. Romero Group Title

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

40. Romero Group Title

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