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Find the areas bounded by the indicated curves, using (a) vertical elements and (b) horizontal elements.$$y=4 x, y=x^{3}$$

Calculus 2 / BC

Chapter 26

Applications of Integration

Section 2

Areas by Integration

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for this problem we are asked to find the areas bounded by the curves, Y equals four X and Y equals X cube. So our first step here is to find their points of intersection. So I will be when four X equals X cubed or when four x minus x cubed equals zero. Or can factor out the X. We'll have then that that will be when X times four minus X cubed equals zero. So that then means that we need to have either X equals zero is probably our left hand point of intersection. And or we need x cubed to equal four. So we need X two equal forward to the power of 1/3 Cube Cubed Root of four. So to begin we need to have Yeah or actually let me just do a quick sketch here. We know we have four X which all just draws a straight line. We have x cubed. So we know that we'll have one point of intersection at 00 and the other point of intersection will be at X equals four to the power of 1/3. I'll write that down here Actually We had four to the power of 1/3 Or cubed root of four. And the corresponding why value would just be why equals for so for part A we're asked to find the area between the curves using vertical elements. So we know that we're integrating from zero up to 4th power of 1/3. Our upper function there is our straight line. So we'll have four x minus X cubed dx integral of four. X would be four X squared over two. So two X squared integral of X cubed will be X power 4/4. We're integrating from zero up to 4th power 1/3. Yeah, so we'll have two times four to the power of 1/3 squared or so. That should be two times four to the power of to over three minus four. To the power of 4/3 over four. So that would just be four to the power of 1/3 which has a value Decimal value of about 3.45. Then we are asked to go through the same process but using horizontal elements instead. Well we were given that Y equals four X. So we need to express this as X equals Y over four and we have Y equals X cube. So that will become X equals Y to the power point over three. And we can see that using horizontal elements. Our upper function, so to speak is going to be that wide power of 1/3 and will be integrating for why between zero and four. So we'll have wide span of 1/3 -Y over four. Dy now integral of white power of 1/3 will be wide span of 4/3 divided by 4/3. So that's times 3/4 Then integral of why over four will become y squared over eight. 1/4 times y squared over two. So you have minus Y squared over eight, evaluating that from zero up to four again will give us a final result equivalent to 2 to the power of or two times 4 to the power of to over three -4 to the power of 1/3, Or approximately 3.45.

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