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Numerade Educator

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Problem 23 Medium Difficulty

Evaluate the integral.

$ \displaystyle \int^0_{-2} \biggl( \frac{1}{2}t^4 + \frac{1}{4}t^3 - t \biggr) \,dt $

Answer

$\frac{21}{5}$

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Video Transcript

We are doing the integral from -2-0. Of this complicated function 1/2 to the 4th plus 1/4 T cubed uh minus T. D. T. So the first thing you need to do is find the anti derivative. And what you do is you add one to each exponents of four plus one is five. And then you multiply by the reciprocal of your new exponents. Well one half times 1/5 is, 1/10 is a multiply straight across the year. Denominators So add one to your exponent and then do 1/4 times 1/4. 1 16. I had one to your exponent And then multiply by the reciprocal of a new excellent. And that's going from negative 2-0 Now. I think it's important, especially if you're new to this, to plug in your upper bounds for all of those but zero to any power still zero. So we're looking at, you know zero, well in 10th of zero is still zero plus zero minus zero. But I think it's important that you write that because if you don't and you only plug in this lower bound and everywhere. Um you might forget that you need to have a minus in front. So negative two to the fifth power. First of all, it stays negative 248 1632. That's how many I was just multiplied by 2.5 times 2. 4. 8. 16 -2 to an even power makes it positive 16, 16 And the -2 squared is positive. Um, positive four half of four would be too. So hopefully that all those signs made sense. Um So again, I'm not going to concern myself any of those zeros we have the negative of, I mean this does reduce to 16 fists And then you can get the same denominator on these sort of be like five fists is the same thing as 166 because it's equal to 1 -10 fists is equal to two. So now you can look at that and say, okay, negative 16 plus five is negative, 11 minus 10 is negative 21 but you have the negative of negative 21 so it's positive 21 fists if you do that in your head.