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In Example 5.1.2 we showed that $ \displaystyle \int^1_0 x^2 \, dx = \frac{1}{3} $. Use this fact and the properties of integrals to evaluate $ \displaystyle \int^1_0 (5 - 6x^2) \, dx $.

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00:58

Frank L.

00:25

Amrita B.

01:22

Stephen H.

Calculus 1 / AB

Chapter 5

Integrals

Section 2

The Definite Integral

Integration

Oregon State University

Harvey Mudd College

University of Nottingham

Idaho State University

Lectures

03:09

In mathematics, precalculu…

31:55

In mathematics, a function…

00:52

In Example 2 in Section 5.…

01:50

Use the properties of inte…

08:19

Evaluate the integrals…

02:02

Evaluate the integrals.

01:05

Evaluate the following int…

04:26

Suppose $\int_{1}^{4} f(x)…

06:00

Use Theorem 5.5 .5 to eval…

06:59

10:18

00:55

Evaluate the definite inte…

Okay in this problem we already know that the definite integral from 0 to 1 of x squared comes out to equal one third. Uh So we want to use that fact and properties of integral to evaluate the definite integral from 0 to 1 of to function five minus six X squared dx. Uh Well there's a couple of properties of integral that we're going to use. Uh The integral of a difference is the difference of the integral. Uh So this is going to equal the integral from 0 to 1 of five DX. Subtract The integral from 0 to 1 of ah six x squared dx. So let's rewrite integral from 0 to 1 Of five GX. Subtract Now the other property of integral is that we're about to use is the integral of a constant times of function is equal to the constant times the integral to function in other words, the integral, six x squared is really six times the integral of x squared. So we're going to take the six out of the integral sign. Um So six is now going to times the integral from 01 of x squared dx. Yeah, okay, when you integrate a constant but the definite integral of a constant is simply that constant Times uh the interval of integration we're integrating between zero and one. So the length of that integral is 1 0 or just one. So at the end of roll of five D x from 0 to 1 is just five Time to lengthen the interval 1 0 which of course will be five times one. And then we are subtracting six times the integral from 0 to 1 of x squared dx. But this was given to us in the original problem. Uh The answer is one third, so six is really just times in a value of one third, So five times 1 0, 1 0 is one. So 5 times one is 5. Subtract six times one third is to, Final answer is three.

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