10.3.3 Area, Integration by Substitution, and Trigonometry Flashcards Preview

AP Calculus AB > 10.3.3 Area, Integration by Substitution, and Trigonometry > Flashcards

Flashcards in 10.3.3 Area, Integration by Substitution, and Trigonometry Deck (8):
1

Area, Integration by Substitution, and Trigonometry

• When finding the area of a region:
1. Sketch the region.
2. Determine how the rectangles will stack.
3. Find where the curves intersect.
4. Set up the integral.
5. Evaluate.

2

note

- The best tip for working area problems is to pay special
attention to the way you choose to stack the rectangles.
- In this example the curves are defined in terms of x. Vertical rectangles stack well for these curves. If the curves were defined in terms of y, horizontal rectangles would probably stack better.
- Make sure that your limits of integration correspond to the variable of integration. If you are integrating with respect to x, then the limits of integration will be x-values.
- Always draw in an arbitrary rectangle. Use that rectangle to determine the dimensions you will use to integrate.
- Once you get the hang of setting up the integral, you will be able to compute the areas of very exotic shapes.
- Keep in mind the different techniques of integration when evaluating the definite integral. Here, the integral can be solved with a simple substitution.
- Remember to convert back to the original variable before you evaluate the integral along the interval.
- Notice that the integration by substitution is performed on an indefinite integral. To work the problem without this step, you would have to express the limits of integration in terms of u.

3

Find the area of the region bound by x=y^3−y and y=1x/3.

8

4

Find the area under the curve from x=0 to x=2 of y=x√x^2+4 .

A=16√2−8/3

5

Find the area of the region bound between the curves x=1−y4^ and y=x/5+1.

A = 18.9

6

Find the area bound between the two curves:
x = y/2 and y = x^ 2 − 4x.

A = 36

7

Find the area of the region bound by x=1−y^2 and x=y^2−1.

8/3

8

Find the area under the curve from x = 0 to x = 1 of y = (x ^4 + x)^5 (4x^ 3 + 1).

A=32/3

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