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The graphs of two functions are shown with the ar…

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Problem 28 Medium Difficulty

Sketch the region enclosed by the given curves and find its area.

$ y = \frac{1}{4}x^2 $ , $ y = 2x^2 $ , $ x + y = 3 $ , $ x \ge 0 $

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Chapter 6

Applications of Integration

Section 1

Areas Between Curves

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Video Transcript

we want to find the area of the enclosed region. Ah, we have the equations y equals 1/4 X squared. That's the wide problem. We have two x squared, which is the narrow problem we have X plus y equals three, which is the question of a line s. So let's label that X plus y equals three and we're only interested in X greater than zero. So we need to figure out what our limits of integration will be. We have this first point of the second point, and this third point, let's label them A B and C A is the intersection of our two problems. To X squared equals 1/4 X squared. So this tells us that X is equal to zero. That's the order. Looking at the second point of intersection we have to solve for two x squared are narrow. Problem equals the equation of her line. So our line is why equals three minus x. So this is just a quick erotic. So solving it gives us two solutions we have one of them is X equals one and the other is X equals negative three over two. But we only care about ex being positive, so we retake X equals one. Next, we're looking at the third point of Intersection, which is going to be the intersection of our wide problem. 1/4 X squared and our line y equals three minus X. Again, it's a quadratic so we can solve it of by using the quadratic formula whereby factoring and what we get is we have the solutions X e quote negative six where X equals two. But since excess positive, we take X equals two so we can label these points along the X axis. This is X equals zero X equals one and X equals two. So we're going to integrate to determine this area. Let's open up a new page. Now here we have area equals into grow work. First, we're going from 0 to 1. So we look at our top function, which is to two x squared over here, and our bottom function is 1/4 X squared. So we're going to subtract those Ah, we have bracket to X squared minus 1/4 X squared DX plus integral. We're going from 1 to 2. And now let's take a look at what is the top function. The top function is the lines. So that's going to be three minus X. And then the bottom is the wide problem. 1/4 X squared. So we have three minus x on. We're subtracting the wide problem minus 1/4 X squared DX. This first sent a girl we can do in one piece because two X squared minus 1/4 X squared is just 7/4 X squared. Ah, so we integrate the anti derivative of 7/4. X squared is seven over 12 x to the three. This is going from 0 to 1 plus. Now we take the anti derivative of this second integral. We take three X minus half X squared. Mmm. And then here we have minus one over 12 X to the three. This is going from 1 to 2 and then after plugging in our numbers, So here we have a seven over 12 minus zero. Next we plug in to hear we have a six minus two squared is four former toos, too, to Cuba's 88 over 12 is three over four. Now we plug in 13 minus 1/2 minus one over 12 and then simplifying everything gives us the solution. Three over two

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Problem

The graphs of two functions are shown with the ar…

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Calculus: Early Transcendentals

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Video Transcript

James Stewart

Calculus: Early Transcendentals

Catherine Ross

Heather Zimmers

Michael Jacobsen

Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Catherine Ross

Heather Zimmers

Michael Jacobsen

Joseph Lentino

James StewartCalculus: Early Transcendentals

Catherine Ross

Heather Zimmers

Michael Jacobsen

Joseph Lentino

James Stewart

Calculus: Early Transcendentals