Area Between Two Curves

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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[@TheOrganicChemistryTutor]

Full 2 Hour Video on Patreon: www.patreon.com/MathScienceTutor Direct Link To The Full Video: bit.ly/3DGVcYy

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[@gelina2650]

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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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. It's quicker, better if your exam is time bound.

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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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