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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 Lin

00:25

Amrita Bhasin

01:22

Stephen Hobbs

Calculus 1 / AB

Chapter 5

Integrals

Section 2

The Definite Integral

Integration

Harvey Mudd College

Baylor University

Idaho State University

Lectures

05:53

In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

40:35

In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

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In Example 5.1.2 we showed…

As Shown.

In Example $5.1 .2$ we sho…

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In Example 2 in Section 5.…

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Use the properties of inte…

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Use the method of substitu…

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Evaluate the integrals.

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Evaluate the integrals usi…

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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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