Problem

Evaluate the integral. $ \displaystyle \int^{…

00:40

Problem 39 Hard Difficulty

Evaluate the integral.

$ \displaystyle \int^{8}_{1} \frac{2 + t}{\sqrt[3]{t^2}} \,dt $

Answer

$\frac{69}{4}$

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Video Transcript

We are looking at the interval from 1 to 8 of this fancy function two Plus T. Over uh the cube root of T. Square Gt. Now I would tell my students that this problem can be easier, you know manipulated because we're dividing by a mono meal that we could just divide each piece by that. So I'm doing this part in the red and remember if you're in the denominator it's really a negative exponents and then the cube root goes into the denominator of the exponents. So um the cube root third power. Okay so then we have plus, I can do the same thing with this and remember you're subtracting the exponents when you're dividing. So 1 -2 3rd. I will give me to the 1/3 power. And now we can do the power rule where we add one to the exponent. So negative 2/3 plus one would be 1/3. And then you multiply by the reciprocal of your new. Excellent. Well two times three because that's a reciprocal of one third is six. And then this would be uh to the 4/3 power uh and multiply by the reciprocal of that would be three force times that From 1- eight. Now I think you should be good enough and you can tell me if I'm wrong, but you should know that when you plug in eight and for both of these bounds that this is the same thing as the cube root of eight, which is two times 26 Or give me 12. And same thing with this, if I plug in eight of the cube root of eight which is two and then to the fourth power is 16. Um and I would probably just simplify like 16 divided by four is four times three will give me 12. And then plug in your lower bound, which is actually kind of nice because one to any power is just Uh one and then times six plus three Force. Um so I would probably do this my head like to get 24 -6. Um So I might even leave it as A mixed number of 18 and 3/4. Except that's wrong because I need to subtract off each piece, So it wouldn't be 18 and 3/4. It would be 17 and 1/4. There we go. But if you change it to an improper fraction, you would do four times 17, Which is 68 plus the one on top. So it'd be 69 force. Um Most people would prefer this answer, but my brain kind of works a little bit differently. They're both correct.

Gregory H.

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 5

Integrals

Section 4

Indefinite Integrals and the Net Change Theorem

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