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# Evaluate the integral.$\displaystyle \int^1_0 (1 - 8v^3 + 16v^7) \,dv$

## $$1$$

Integrals

Integration

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

Lectures

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### Video Transcript

Okay, let's go ahead and do this problem were given the integral off one minus eight V Cube plus 16 V to the seventh TV. So in order to evaluate this expression, all we need to do is to utilize thes two facts, the power rule off the, um not the derivative, the anti derivative of the integral and the second part off fundamental team of calculus. So let's remember this portion V to the power. When you take the derivative, you will get envy to the N minus one. So the power is reduced by one and then you multiply and right, So because this is an anti derivative, you go in the opposite direction. The first thing that happens is that the power is increased by one. And then instead of multiplying the number that you had in the exponents, now you're gonna be dividing by the number that you have in the Expo. So this is the expression that you're going to get next. The fundamental theme of calculus basically says once you find the anti derivative, all you have to do is to plug in the B and the A in the way that it's going to be a subtraction off the anti derivatives. So this is a very powerful fact that makes all of the calculations easy. So let's apply to this problem. The derivative theme integral off one is going to be V the Cube. The integral would be V to the fourth divided by four. So you will have minus eight V to the fourth over four. Next expression is going to be 16 v to the eighth over eight, and then you're going to evaluate it by plugging in the numbers zero and one later. Now, when I plug in zero, you can see that all of these expressions, they're going to be zero. So I am not going to worry about it. And plugging in one is also quite simple. So you just need to say this is one. A divided by four is too, and 16/8 is also to these two terms. Cancel each other out. So the result is one, and this is how you solve these kind of problems.

University of California, Berkeley

#### Topics

Integrals

Integration

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

Lectures

Join Bootcamp