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Use the Midpoint Rule with the given value of $ n…

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Problem 8 Easy Difficulty

The table gives the values of a function obtained from an experiment. Use them to estimate $ \displaystyle \int^9_3 f(x)\, dx $ using three equal subintervals with (a) right endpoints, (b) left endpoints, and (c) midpoints. If the function is known to be an increasing function, can you say whether your estimates are less than or greater than the exact value of the integral?


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Frank Lin

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Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 5

Integrals

Section 2

The Definite Integral

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Integrals

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Video Transcript

All right. We've got a question here where we're giving a table of values and we want to estimate the Integral from 3- 9. The function of X With respect to X. And we're gonna use three equals sub intervals. And we want to calculate for a the right The the left hand points and see the mid points. And then finally, if it is a function is known to be an increase in function to say whether you're estimates or less Then operator than the exact value. Alright. So to start off a part a, What we're doing is we know we have that three sub inter rules 3 2579. And we're What we're asked to do is use your right endpoints for part a. So to begin, we know we're going to have three sub intervals. So the way that we would calculate for our change in X is you take your largest value within that interval And you subtract us lowest value within that Um interval. And then you divided by your southern rules that was given to us as 3. And you get to as your change in X. Okay. And then finally the way that you saw for your essentially your dream and some on the right end is you take your change in X which is two and then you multiply by the function of X. When you are functional X one was the function of except to What's the function of except Because we have three sub animals. All right. So you start off by saying, well what would be your X sub one? Your X one would fall on the the first Sub in a role. Excuse me, let me let me reiterate that to be a little quicker. So you have your your range of sub intervals we're told or three. Right. So you would have This being your one in some interval, this being your second interval and that's being a right, that is an equal space between three and nine. And then you have created a 3 subgroups. Okay, so whenever you're asked to solve for your Right and points, you take your first interval and choose the value on the right end. If you can see that our um interval, the first intervals from 3 to 5, second intervals from 5 to 95 to seven and then seven and nine. So you would choose the value on the right hand side. So you would say well your ex a one in five then we know that X when x one is five, then you have a negative six. Yeah, The X up to it's 0.3 Excuse me. Except to would be at seven because that's the right end on your second sub interval And that is when the function of that's easy, quick to point out, Okay, and then your exit three is nine because that is the right endpoint on your third interval and that is when the function that's at one point. All right. So you're basically gonna be left with a final answer which is 4.2, alright, for part B were asked to the same thing, but this time on the left end points you take your same intervals and your instead of X of one being um five, you really would start off by saying except zero except one and exit to Basically what you're doing is you're choosing the left endpoint On your interval, which here would be three. We know when X is equal to three, we have a negative 3.4 In the left endpoint on her second sub in a rule It's gonna be when you have five and that's at negative .6. Finally, the left endpoint on your third. Some interval here is when X is equal to seven and that is point now. Okay. Your final answer will come out to be -6.2. All right. Finally for port, see taking your um Mid points of your intervals, you can see the midpoint would be between three and five which is four. We know an exit sequence of four. You have a negative 2.1 If your second sub interval is between five and seven, so the midpoint would be at six and that's .3. Finally in the third sub interval value in the middle here is the main point is at X is equal to eight, which would be the function that's equal to 1.4. All right. So then when you go ahead and multiply orders across your left with a point negative 0.8. Alright. And those will be all of your final answers there. I hope that clarifies the question. Thank you so much for watching.

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40:35

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In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

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