Areas+Between+Curves


 * //__Areas Between Curves__//**

To find the area between curves we set up definite integrals solve them using the fundamental theorem of calculus.


 * Example 1**: Find algebraically the area of the region in quadrant 1 that is bounded by the graphs y=2x and y=[[image:http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=x%5E2]] and the y-axis.

First, graph or plot the graphs on you calculator or on paper. Identify the region that the problem asks for.



__Figure 1-1a__

Now draw a strip with a width dx inside the region. Since dx is small, the length of the strip isn't very different on the left from the right side.

__Figure 1-1b__

Now, we see that as dx approaches zero, the lengths become equal. Because of this, the length of the strip is simply. Since the strip is technically still a rectangle, its area can be found with the expression. Let dA be the area of the strip. Just remember always to subtract the larger y from the lower y so you get the positive value for the height of the rectangle. Now, before integrating, you find the limits of integration. The limits in this case are x = 0 and x = 2, as can be seen in the graph. If you cannot tell by looking at the graph, the points can be found by setting the two functions equal to one another and solving for x.

2x =

x = 0 and x = 2

Now note:

A ≈ Σ dA = Σ (2x - ) dx

A = (2x - ) dx = ( + c) - I =  + c - /3 - c] I =  - /3 I  = 4 - 8/3 - 0 - 0 = 4/3

Sometimes, vertical slicing makes the problem hard to solve. In this case, use horizontal slicing and use the same method. When you subtract them in the integrand, subtract the rightmost function from the leftmost function.


 * Example 2**: For the region bounded by the graphs of x = [[image:http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=y%5E2]] - 3 and x = y + 3, write an integral for the area and evaluate it numerically. Evaluate the integral exactly by the fundamental theorem and compare your answers.

First, pick points (, y) and (, y).

Then:

dA = dY = ( - y - 6) dY

The graphs intersect where the two x values are equal:

- 3 = y + 3

y = -2 and y = 3

The total area is found by adding all the dA's and taking he limit A.K.A. integrating. Also, be sure that the limits of integration are the y values!

dA = ( - y - 6) dY

A = ( - y - 6) dY

Using numerical integration:

A = -26.166

Using Fundamental Theorem:

A = /3 - /2 - 6y I {3, -2}

A = (27/3) - (9/2) - 18 - [(8/3) - (4/2) - (-12)]

A = 9 - 4.5 - 18 - 2.666 + 2 - 12 = -26.16

Here's a video similar to ex. 1:

media type="youtube" key="DRFyNHdVgUA" width="425" height="350"

Here is a problem similar to ex. 2:

media type="youtube" key="70NQ3ISYihw" width="425" height="350"

...and here's part 2:

media type="youtube" key="Xne6Hv9useE" width="425" height="350"

media type="custom" key="6091633"

Video 1: http://www.youtube.com/watch?v=DRFyNHdVgUA&feature=player_embedded

Video 2: http://www.youtube.com/watch?v=70NQ3ISYihw&feature=player_embedded

Video 3: http://www.youtube.com/watch?v=Xne6Hv9useE&feature=player_embedded