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Problem 70 Hard Difficulty
(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
$$ \biggl| \int^{2\pi}_0 f(x) \sin 2x \,dx \biggr| \le \int^{2\pi}_0 \bigl| f(x) \bigr| \,dx $$
Answer
(a) $\left|\int_{a}^{b} f(x) d x\right| \leq \int_{a}^{b}|f(x)| d x |$
(b) $\left|\int_{0}^{2 \pi} f(x) \sin 2 x d x\right| \leq \int_{0}^{2 \pi}|f(x)| d x$
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Top Calculus 1 / AB Educators
Catherine R.
Missouri State University
Heather Z.
Oregon State University
Kayleah T.
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Watch More Solved Questions in Chapter 5
Video Transcript
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.
Top Calculus 1 / AB Educators
Catherine R.
Missouri State University
Heather Z.
Oregon State University
Kayleah T.
Harvey Mudd College
Michael J.
Idaho State University