14–25. {Use of Tech} Areas of regions Determine the area of th...

Question

14–25. {Use of Tech} Areas of regions Determine the area of the given region.

The region bounded by y = x²,y = 2x²−4x, and y = 0

Accepted Answer

Welcome back, everyone. In this problem, we want to determine the area of the region bounded by Y equals X2 + 2 X, Y equals 2 X2, and Y equals 0. A says it's 8/3 square units, B 3/8, C 4/3, and D 3/4 square units. Now, before we can find the area of the region bounded by our curve, let's start by finding the points where the curves intersect. That will give us the interval at which the area, or sorry, the interval of our region. Now, at the points of intersection, OK. Then the two curves will be equal to each other. In other words, X2 + 2 X. K is going to be equal to 2 X2. Now if we subtract 2 X2d from both sides, then we get negative X2 + 2X to be equal to 0, and then we can factor out X from our left hand side to get X multiplied by negative X + 2 to be equal to 0. In that case, X will be equal to 0, and negative X + 2 will be equal to 0, which means X is going to be equal to 2. So the curves intersect at X equals 0 and X equals 2. No, that can give us a better idea of how graph is to look because if we were to go ahead and try to sketch or graph to show our shaded region here, now, starting with Y equals X2 + 2X in red, this is going to be a parabola opening upward. Its vertex is going to be a -1, -1, OK. So that means if I were to just label my axis here, the graph of Y equals X2 + 2 X would probably look something like this, OK? Because we know that it passes through the 000. It passes through the origin. Now if we move on to our next graph of Y equals 2 X squad in blue, we know that it will also be a parabela, opening upward but steeper than the first because it's the coefficient of Xsquared is larger and we also know its vertex is going to be at the origin. So that graph will probably look something like this, OK? All right, let me draw it a bit steeper here, probably something like this. Now, we know, like we said, they intersect at X equals 0 and at X equals 2. So our region is this section shaded in black, OK? So basically, this is what we're trying to find. Now how does this graph help us? Well, this can help us to figure out which function is on top so we can know how to structure them in the integra. And from the graph we can see that Y equals X2 + 2 X is above Ye equals 2x2 on this interval. So to find the area, OK. Between the curves. From X equals 0 to X equals 2. We can find our definite integral between the limits of 0 and 2 of X2 + 2 X. Minus 2 x squad all with respect to x. Now if we expand our term here. This is going to be the definite integral of negative X2 + 2 X with respect to X, and now we can integrate by integrating each of these terms separately. The integral of negative X squared will be a third of negative Xcubed, and the integral of 2 X is equal to x2. So we have a third of negative X cubed plus x2 between the limits of 0 and 2. To solve, we can substitute our upper and lower limits. So for our upper limit, this will be equal to -2 cubed divided by 3 plus 2 squad. And for our lower limit we would substitute 0 into both terms which would leave us with 0. Now -2 cubed divided by 3, that's negative 8/3, 2 squared is 4, and negative 8/3 plus 4 is equal to 4/3. Thus, the area of the region bounded by the curves would be equal to 4/3 square units. If we look back at our answer choices, that means C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

14–25. {Use of Tech} Areas of regions Determine the area of the given region.

The region bounded by y = x²,y = 2x²−4x, and y = 0

Key Concepts

Finding Points of Intersection

Setting Up Definite Integrals for Area

Understanding the Role of the x-axis (y=0)

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