AP Calculus AB: 10.3.3 Area, Integration by Substitution, and Trigonometry
This set of flashcards reviews techniques for finding the area between curves using definite integrals, with emphasis on integration by substitution and proper stacking of rectangles. It includes tips on sketching regions, setting up integrals, and handling problems involving trigonometric and substitution methods.
Area, Integration by Substitution, and Trigonometry
When finding the area of a region:
Sketch the region.
Determine how the rectangles will stack.
Find where the curves intersect.
Set up the integral.
Evaluate.
Key Terms
Area, Integration by Substitution, and Trigonometry
When finding the area of a region:
Sketch the region.
Determine how the rectangles will stack.
Find whe...
note
The best tip for working area problems is to pay special
attention to the way you choose to stack the rectangles.In this...
Find the area of the region bound by x=y^3−y and y=1x/3.
8
Find the area under the curve from x=0 to x=2 of y=x√x^2+4 .
A=16√2−8/3
Find the area of the region bound between the curves x=1−y4^ and y=x/5+1.
A = 18.9
Find the area bound between the two curves:
x = y/2 and y = x^ 2 − 4x.
A = 36
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| Term | Definition |
|---|---|
Area, Integration by Substitution, and Trigonometry | When finding the area of a region:
|
note |
|
Find the area of the region bound by x=y^3−y and y=1x/3. | 8 |
Find the area under the curve from x=0 to x=2 of y=x√x^2+4 . | A=16√2−8/3 |
Find the area of the region bound between the curves x=1−y4^ and y=x/5+1. | A = 18.9 |
Find the area bound between the two curves: x = y/2 and y = x^ 2 − 4x. | A = 36 |
Find the area of the region bound by x=1−y^2 and x=y^2−1. | 8/3 |
Find the area under the curve from x = 0 to x = 1 of y = (x ^4 + x)^5 (4x^ 3 + 1). | A=32/3 |