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Make the given changes in the indicated examplesof this section and then find the resulting areas.In Example $1,$ change $x=2$ to $x=3$

$$\frac{26}{3}$$

Calculus 2 / BC

Chapter 26

Applications of Integration

Section 2

Areas by Integration

Missouri State University

Campbell University

University of Michigan - Ann Arbor

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for this problem we are asked to find the area under the plot of Y equals X squared for between X equals one and three. So our first step here is to substitute in Y equals X squared into our integral. So we have the integral from 1 to 3 of x squared dx. Now we should know that using our rule for anti differentiation, the anti derivative of X power of N. Let me write this down properly here. Anti derivative of X power of N. It's going to be exposure of N plus one divided by N plus one plus a constant. So applying that here we have that this is going to be X cubed over three. And we want to evaluate this From 1- three. So first we'll have nine cubed or we'd have three cubed over three or three squared which would be nine And then we would have -1 cubed over three. And so well we need to multiply that back in nine times three is 27. So we have 27 minus 1/3 which would be 26/3 as our final answer for the area.

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