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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.$$f(x)=\tan ^{-1} x ;[a, b]=[0,1]$$

a. 0.62452b. 0.51569c. 0.477677

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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in this problem. We're working with the function, the inverse tangents of X, and we're working on the interval from 0 to 1 to find the area under the curve, I'm going to be using two rectangles, five rectangles and 10 rectangles. We can continue this process for more and more rectangles. The more rectangles will produce a more precise solution, since my interval goes from 0 to 1 that has a total distance of one unit in order calculate the area under the curve and take the with times the sum of all of the outputs at those intervals. So in this case, my total distance is one divided into two pieces means that each rectangle will have a width of one half if I only have two rectangles, each rectangle level with with of 1/5 if I have five rectangles and each wreck were taken level with of 1/10 if I have 10 rectangles when I'm all for the with times the sum of all of the I'll put values for the total number of rectangles in this case too. In this case, I want to take 1 to 2. 1 to 5 and 1 to 10 because that's the total number of rectangles that I'm going to need. What I'm going to have to do is I am going to have to take the inverse tangent and I'm gonna do the inverse tangent, all of the haves. So x sorry, not x and over to. And when I plug that into my calculator, it will give me a result of an approximate area of 0.6245 Now, if I take the inverse tangent of all the fifths at all the fifth units, I'm going to get an approximate area of 0.51 57 And if I take the inverse tangent all the 10th, I'm gonna get an area of 0.4777 I can repeat this for 100 1000 a million, but more the more rectangles I use, the closer might answer gets to a more precise, more accurate answer of the area under the curve. So if this was 100 this would be 100 100 and 100

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