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(a) If $ f $ is continuous on $ [a, b] $, show that$$ \biggl| \int^b_a f(x) \,dx \biggr| \le \int^b_a \bigl| f(x) \bigr| \,dx $$[$ Hint: $ $ -\bigl| f(x) \bigr| \le f(x) \le \bigl| f(x) \bigr|. $]

(b) Use the result of part (a) to show that $$ \biggl| \int^{2\pi}_0 f(x) \sin 2x \,dx \biggr| \le \int^{2\pi}_0 \bigl| f(x) \bigr| \,dx $$

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Calculus 1 / AB

Chapter 5

Integrals

Section 2

The Definite Integral

Integration

Campbell University

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

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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(a) If $f$ is continuous o…

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(a) Let $f$ be continuous …

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(a) If is continuous, prov…

hey will try to prove this person equality and ah, he'LL notice one thing. This absolute value ofthe f works. It's glittery. Who's wherefore? Fax is also greater or equal to next the Fairfax, but by the defendant. Absolute value is a bigger while, these two. So this world tells us that it this integral, it's going to be greater or equal to this integral. And this means this the same into group. It's going to be quitting or equal to this integral for them by linear poverty off the integral. Your factor. All of that, the negative side. So what does it mean? It means thiss quantity is greater or equal to post this one and negative off this one. Let me it's going to go to this one and also the connective off the same quantity. If, ah, if something is greater or equal to this and negative all the same thing, then it's quite a record to the absolute value, because ever so very with the bigger one off these two. But ice greater because of both. So this proof of the first part there's second part fof excited to works by the first part. It's great very for two. Ah the integral. By putting the absolute value inside and this equals and because this's less or equal to one, no sign of anything is between one and one. And if you put the absolute values between zero and the one, that means if we drop it, it's going to make the an integral, bigger and therefore the integral value picker, and that's the second part we need to prove.

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