anonymous
  • anonymous
How can I solve the following? (attached)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
anonymous
  • anonymous
Sample problem:
anonymous
  • anonymous
What step are you confused on the sample problem?

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More answers

anonymous
  • anonymous
I understand the sample problem. I just don't know how to use it for the actual problem.
anonymous
  • anonymous
do the same...instead of using 7 use 5
anonymous
  • anonymous
I did, but it shows up as wrong when I put it in.
anonymous
  • anonymous
then you did something wrong. ...show me the steps
anonymous
  • anonymous
what is the final answer that you got when you did the problem?
anonymous
  • anonymous
Nevermind, I found my error, instead of 2/9 I put 2/5. Sorry. Can you help me with another problem though?
anonymous
  • 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
  • anonymous
anonymous
  • anonymous
anonymous
  • 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
  • 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
  • anonymous
You just integrate what's inside first....can you tell me what you get?
anonymous
  • anonymous
3x^2+7x+C
anonymous
  • 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
  • anonymous
you have to do the chain rule.
anonymous
  • anonymous
Try it one more time....what is the integral?
anonymous
  • anonymous
I have no idea here..
anonymous
  • 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
  • anonymous
Would it be: 1/12 root of 6x+7?
anonymous
  • anonymous
Is the answer 1/4? If yes, then I've done my work correct.
anonymous
  • 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
  • anonymous
Yes
anonymous
  • anonymous
it's suppose to be\[\frac{1}{6}*2=\frac{2}{3}=\frac{1}{3}\]
anonymous
  • anonymous
1/3 root of 6x+7 not 1/12 root of 6x+7
anonymous
  • anonymous
Oh, so then the answer would be: 1.
anonymous
  • anonymous
what no I got something else
anonymous
  • anonymous
\[\frac{1}{3}\sqrt{6x+7}\]\[\frac{1}{3}\sqrt{6(7)+7}-\frac{1}{3}\sqrt{6(3)+7}\]
anonymous
  • anonymous
\[\frac{1}{3}(\sqrt{6(7)+7}-\sqrt{6(3)+7})\]\[\frac{1}{3}(\sqrt{42+7}-\sqrt{18+7})\]
anonymous
  • anonymous
can you do the rest?
anonymous
  • anonymous
Yes, I got it.
anonymous
  • anonymous
what did you get?
anonymous
  • anonymous
Im doing the math right now.
anonymous
  • anonymous
ok tell me what you get just to check if you got it right.
anonymous
  • anonymous
2/3
anonymous
  • anonymous
yeah that's what I got good job.
anonymous
  • anonymous
If you have more problems post them in a different question.
anonymous
  • anonymous
I want to give you more medals :) for the good job

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