How can I solve the following? (attached)

- anonymous

How can I solve the following? (attached)

- schrodinger

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- anonymous

##### 1 Attachment

- anonymous

Sample problem:

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- anonymous

What step are you confused on the sample problem?

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## More answers

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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- anonymous

Sample:

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- anonymous

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

- anonymous

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
\[\int\limits_{3}^{7}(6x+7)^{-\frac{1}{2}}dx\]

- anonymous

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

- anonymous

3x^2+7x+C

- anonymous

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

- anonymous

you have to do the chain rule.

- anonymous

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

- anonymous

I have no idea here..

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Yes

- anonymous

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

- anonymous

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

- anonymous

Oh, so then the answer would be: 1.

- anonymous

what no I got something else

- anonymous

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

- anonymous

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

- anonymous

can you do the rest?

- anonymous

Yes, I got it.

- anonymous

what did you get?

- anonymous

Im doing the math right now.

- anonymous

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

- anonymous

2/3

- anonymous

yeah that's what I got good job.

- anonymous

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

- anonymous

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

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