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Find the areas bounded by the indicated curves, using (a) vertical elements and (b) horizontal elements.$$y=x^{3}, x=0, y=3$$

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 area bounded by the curves. Y equals x cubed, X equals zero and y equals three. The first step that I'm going to take here is to just create a rough little sketch of our curves. So I'll have first the Y equals X cubed. We'll do in the blue here, you know that that looks something like this, then I'll do the X equals zero that corresponds to the y axis. And then also we want along when Y equals three. So it'll be something over here. So we want to find the area inside of this. So for the first part of the problem we are asked to do this using uh using vertical elements. So we'll need to figure out where the X or what the X value is when y equals three. So we can figure out what to integrate X over. Well if we have that, Y equals X cubed and Y equals three. So we would have that three equals X cubed And so X is going to equal three to the power of 1/3. So To find the area will take the integral from zero 2, 3 to the power of 1/3 of now. Our upper function or upper line is going to be three. So it will be three minus are lower line, it will be three minus X cubed dx. Now integrating three will give us three X and integrating negative X cubed will give us x power of 4/4 Evaluating that from zero up to three speaker, 1/3. We can express this in the form, the most simplified form. Uh Nine times three to the power of 1/3 divided by four. Now for part B. And I'll note that the numerical value there is approximately 3.245 now for part B were asked to do the same but using um horizontal elements instead. So what we first need to do is we know why is a function of X means then we have to translate this into X as a function of why? So we'll have x equals white power of 1/3 is our function and we are going from y equals zero up to y equals three. So The integral will be from 0 to 3 of why? It's the power of 1/3 D. Y. Well I'll note that the lower bound on the function would be the line corresponding to X equals zero or y equals zero. So anti derivative of this would be wide to the power of 1/3 plus one. To be white power for over three divided by 4/3. So it's the same thing as multiplying by 3/4. We're evaluating that from zero up to three. So we can see that we'll get three times three to the power of 4/3 over four which is equivalent to nine times three to the power of 1/3 over four. And so we can see that we have the same result either way.

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