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Problem 5 Medium Difficulty

Evaluate the line integral, where $ C $ is the given curve.

$ \displaystyle \int_C (x^2y + \sin x) \, dy $,
$ C $ is the arc of the parabola $ y = x^2 $ from $ (0, 0) $ to $ (\pi, \pi^2) $.

Answer

$\frac{1}{3} \pi^{6}+2 \pi$

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Video Transcript

So we do this line to grow here we have the Y and as usual, the first step is to Parametric is, uh, the ark. So X equals T y whose T square and as we can see from here, he clears clearly goes from zero to pi So the integral will be zero to pi X square Y should be teaching the four plus sign Why is these others X s t so X square was cheating the fourth and access key and the wife is to t d g duty to the fifth plus two t 70 d p Mhm. Okay, so let's try to, um, try to compute the integral this one separately or if you remember, the format is always good. It will have to I'll use integration by part. We said this to be you and is to be t. V. And this just gives you you times feet which should be negative t co society, uh, minus fee which is co sign t the U D t minus and VVS negative cosign t So we should have plus here. So this one we integrate with our scientists. We are scientific minus t coz I T plus C and I hope this is correct. Let's check it quickly. By taking the derivative we got a call sign T here. We got a by product rule coz I shit minus t times negative scientists plus t sign here. That's right. Uh huh. We do have this anti derivative formula is open another page who want zero to pi to t fifth plus two t 70 d p and we write out the anti derivative. So it's teachers by power roll This should be t to the six over six times too. So over three plus two times formula we find also to side T minus two ti ko 70 We're probably in pie. We got pie to the 6/3 plus cyp I zero and the cold sign pie is negative one. So here we got zero and the minus two pi times that he wants a plus two pipes and the miners. If we probably zero first term is zero second terms. Every term is zero. Actually, you know, this is 0700 and we have a zero here, so So we don't really be this term. And then this is the final answer