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

0.85,0.75,0.71

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 were given the function why F of X is equal to the square root of X on the interval from 0 to 1, we want to estimate the area underneath the curve. And since I'll be doing this by hand of focus on the smaller values. If you have a calculator that you can do this on, you can use higher values will get a more precise answer. So in this particular case, what we're going to do is the first thing we're gonna do is we're gonna break this up into two rectangles in order. Break this up into two rectangles. What we're going to do is we're going to take the squarer of one over to the denominator is the end value plus the square of 2/2. So these are my two triangles, times my change in X right, which is one over to some cutting it and a half. All right, so in this case, what we have here is when I enter this into my calculator, I'm going to get an approximate value here of 0.85 36 And this isn't a private area approximation with two rectangles I can repeat this process for five rectangles as well. This will be the square of 1/5, plus the square of 2/5 plus two squared of 3/5, plus the square of 4/5. Close the square of 5/5 and then multiply that by 1/5 when I'm out. When I put that into my calculator, I'm going to end up with 0.74 97 I can repeat this process for 10 as well. The square of 1/10 squared of 2/10 plus the square root of three tense, close to square to four tense that that that that that plus squared of 10 tents almost wide by 1 10 And so when I most by that, put that in my calculator, I am going to get I'm entering into my calculator as we speak. One tent and then chew tense and then three tense, then four tense on 5/10 plus six tense plus seven tense plus eight tense plus squared of nine tense plus the square of 20th, and I should get an approximate value of 0.7105 The more intervals that I get the closer that this will become

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