Area Between Two Curves
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- Опубліковано 18 кві 2021
- This calculus video tutorial provides a basic introduction in finding the area between two curves with respect to y and with respect to x. It explains how to set up the definite integral to calculate the area of the shaded region bounded by the two curves. In order to find the points of intersection, you need to set the two curves equal to each other and solve for x or y. You need to be familiar with some basic integration techniques for this lesson. This video contains plenty of examples and practice problems.
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Good
Thx bro 😅😊
0:14 = Area from a to b (vertical strip)
2:25 = Area from c to d (horizontal strip)
Example problems:
4:20 (1) Calculate the area bounded by the region by the line y=8-2x, the x-axis, and the y-axis
9:41 (2) Calculate the area of the region bounded by the line y = x and y=x^2
13:25 (3) Calculate the area of the region bounded by the curves y = x^2 and x = y^2
19:40 (4) Calculate the area of the region bounded by the curves x=1-y^2 and x=y^2-1.
24:51 (5) Calculate the area of the region bounded by the line y = x^2-4x and the x-axis.
31:22 (6) Calculate the area of the region bounded by the equations y=x^2-4x and y = 6 - 3x.
39:35 (7) Calculate the area of the region bounded by the equations x= 3y-2
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Professor Organic Chemistry Tutor, thank you for a fantastic video/lecture on finding the Area Between Two Curves in Calculus Two. Graphing the function/functions is very important when finding the Areas Between two Curves in Calculus. All the examples are off the learning charts in this great video. This is an error free video/lecture on UA-cam TV with the Organic Chemistry Tutor.
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First you need to know the concept of limits, then you need to know how to take the derivatives of functions, then you need to know how to do the reverse (anti derivatives), then with anti derivatives you can find integrals and that’s the only calculus part in this specific lesson
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For number 4 at 17:40
Once you know the shape of the curve of x=1-y^2, you can just rotate it 90° and calculate the area of y=1-x^2 from -1 to 1, after that multiply it by 2 (because x=y^2-1 is mirroring x=1-y^2) and you get the same answer, just alot faster if you're used to x-curve calculations.
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What area are you solving for if you subtract the grater function from the lower function? Or in the y-axis, the righter function from the lefter function?
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Hey, thanks a lot for the videos and explanations. I got a doubt, as in the 5th exercise. By convention and notation, this area falls under the x-axis. Must the result be expressed as a negative number, supposing that's "the area above the graph"?
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제가 흔히 봐왔던 그래프는 y가 종속변수이고 x가 독립변수인 꼴의 그래프였습니다. 공부를 함에 있어서 점점 많은 형식들이 이러한 종속변수와 독립변수의 개념이 모호해지고 그러한 적분을 수행해야할 때가 많았는데 그러한 적분방법의 기초를 이 영상을 통해 다시 상기할 수 있었습니다.
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15:22
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Ple solving determine the area below f(x)=3+2x-x^2 about x axis.
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