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Estimate the area between the graph of the function $f$ and the interval $[a, b] .$ Use an approximation scheme with $n$ rectangles similar to our treatment of $f(x)=x^{2}$ in this section. If your calculating utility will perform automatic summations, estimate the specified area using $n=10,50,$ and 100 rectangles. Otherwise, estimate this area using $n=2,5,$ and 10 rectangles.$$f(x)=e^{x} ;[a, b]=[-1,1]$$

3.71828, 2.85174, $2 \cdot 59327$

Calculus 1 / AB

Calculus 2 / BC

Chapter 5

INTEGRATION

Section 1

An Overview of the Area Problem

Functions

Limits

Differentiation

Integrals

Integration

Integration Techniques

Continuous Functions

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So this problem we're working with F of X is equal to e to the X power on the interval from negative 1 to 1. This tells us that my interval goes from negative Oneto one which a grand total of two units we're using wreck tank, two rectangles, five rectangles and 10 rectangles in order Approximate the value. Now the procedure will continue. We can use all of the many rectangles as we want. The more rectangles we used, a more precise my answer will be So we need the wits of all of these rectangles. Well, if the distance is too and I'm breaking a two rectangles, might with will be one to over to If my with is two and a break up into five rectangles, each with will be 2/5 And if I have a with a total distance of Jew and I have 10 rectangles, it's gonna be 1/5 or too tense. I'm gonna multi the with times the sum of all of the heights and that will give me the approximate value for each of the for each of the sums. So in this case, I'm gonna do to rectangles, So I have from 1 to 2 and I hav e And again my starting value is negative. One plus all of my with. So we've got my two x over twos, and when I substitute that into my calculator, I'm going to get 3.71 83 and this is an approximation of the area under the curve. Now, when I switch this up for five rectangles, the formula is not gonna change very much. But the choose that represent the number of rectangles will change. So in this case, I have the some from end goes toe 1 to 5 in five rectangles e from negative one plus. Now, I'm gonna do all the 2/5. So two X over five. When I get my approximation here, I'm going to have about 2.8517 I'm gonna repeat that process one more time and again I can keep repeating it for more and more rectangles. If I wanted to become war and more precise in this case, I'm gonna replace all those fives with tends. And in this case, I have the some from one goes to 10 meeting 10 rectangles e to the negative one plus all the two tents, two X over tens, and that's gonna give me an approximate value of 2.593 three. The more rectangles I use, the more precise my answer will be.

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