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Determine the area of the region between the given curves.$$f(x)=2 x^{3}-3 x^{2}+2 x-1 \text { and } g(x)=5 x^{2}-4 x-1$$

$37 / 6$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 8

Applications of the Definite Integral

Integrals

Campbell University

Oregon State University

University of Michigan - Ann Arbor

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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Determine the area of the …

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01:45

Find the area of the regio…

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01:42

for this problem we are asked to determine the area of the region. Between the curves. F of x equals two X cubed minus three, X squared plus two X minus one and G of X equals five X squared minus four X minus one. So the first thing that we want to do here is solve for when these two curves will cross. So we want to x cubed minus three X squared Plus two x -1 minus five, X squared plus four X +12 equals zero. So that would be two x cubed than minus eight, X squared plus six. X must equal zero. We can see that we have a common factor of two X between all of our terms. So we can write this as two x times X squared minus minus four X plus three. Now this will have a solution when X equals one. Or excuse me when X equals zero. Then from the second pair of brackets there We would have solutions at x equals one And x equals three. So we would have that the two lines are going to cross each other at those three points. And we'd have we'll just create a quick sketch here. I'll use red first so we'll have our parabolic function and then we have the cubic function and I'm just going to be sort of roughly sketching this. They're gonna equal each other when x equals zero. Then the parabolic function is going to shoot up above but then the cubic function, it doesn't go backwards but that's just an attempt here um the cubic function comes back up and goes meets with the parabolic function again. So the area between the two curves is going to be this area here. Plus this area here. The significance is that we need to make sure that we identify which function is the greater one correctly. So We'll have that the area between the two curves will be the integral from 0 to 1 of the cubic function minus the parabolic function. So that will be what I already wrote up above here. Actually that was the cubic function minus the parabolic. It would be two X cubed minus eight X squared plus six X six x dx plus the integral From 1 to 3 of the negative of that. So it be negative two x cubed -8 x squared plus six x dx. So we would then have that a vote integrating our first or executing our first integral there. We'd have two times X power for over four. So it would be X The power of 4/2 minus eight X squared over. Or sorry, eight x cubed over three. So it'll be minus 8/3 X cubed plus six X squared over two. So it'll be plus three X squared Evaluated from 0 to 1. Then we would have minus, you know, put this in brackets here. Well actually no, I'm going to be careful here so we can bring out that minus From the set of brackets that I indicated for that. 2nd integral. So we're subtracting off the same integral but evaluated from 1-3. So that would then be minus X. Power 4/2 -8/3 x cubed plus three X squared, Evaluated from 1 to 3 this time. So for our first evaluation we'd have 1/2 minus 8/3 plus one. Alright, Excuse Me Plus three. Then we'd have minus 3 to the power 4/2 -8/3 times three cubed. So that would be -8 times three squared. So minus eight times nine plus three times three squared. So plus three cubed. Then -1/2 uh minus or a plus 8/3 minus three. So we'll note that from evaluating at the bottom end of our second integral, we'll add up to zero with all the terms from our first integral. So we just need to evaluate what we have in here. Excuse me, I need to correct myself here. Made a mistake involving the fact that we would have double negatives. So actually those terms will add up constructively. So we have that the 1/2 minus 8/3 plus three is going to be 5/6. So we have two times five or six or 5/3. Then the other term remaining there from evaluating the second integral at that at the point X equals three would be plus nine over to So this is going to evaluate out to 37/6.

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