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Repeat Exercise 47 for the curve $ y = (x^2 + 1)^{-1} - x^4 $.

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00:32

Frank Lin

01:28

Amrita Bhasin

Calculus 1 / AB

Chapter 5

Integrals

Section 4

Indefinite Integrals and the Net Change Theorem

Integration

Missouri State University

Campbell University

Oregon State University

Boston College

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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Alright based off of the problem, I'm trusting that are allowed to use a graphing calculator plus the equation is kind of awkward. X squared plus one to the negative first power. And then we have minus X. To the fourth hole account. I don't know how you would do this in your head and maybe there's some fancy math going on. But once you do that you can zoom in and we actually only care about the X coordinates. This must be an even function because it looks symmetric with the Y. Axis. And we care about negative .8692 positive .869. So that's what I'm gonna do. I'm gonna pull up my tool under my functions and sometimes you have to click around before you find where your integral is. Mhm. And I believe I typed that I remember that. So negative .8692 positive .869. Yeah. And uh you unfortunately I might have to re type this equation otherwise you can copy and paste it down. Um And then you also might want to tell the calculator that X. Is your independent variable. I'm sure some other calculators might do that for you. Um and I rounded these answers these bounds, so my answer might be off a little bit, but if I were rounding my answer I would write down 1.233. Rounding to three decim, otherwise you can copy down the whole thing. Um And it should make sense that that answer is correct because if you look at trying to zoom around and it's not letting me and alright, there we go. Uh That that looks about right, the area looks like it should be a little bit bigger than one but smaller than two. Um things like that. So 1.233 is what I would write down.

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