 # Drill Find the area between the x-axis and the graph of the function over the given interval: y = sinx over [0, π] y = 4x-x 3 over [0,3]

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Drill Find the area between the x-axis and the graph of the function over the given interval: y = sinx over [0, π] y = 4x-x 3 over [0,3]

7.2: Applications of Definite Integrals Day #1 Homework: page 395-396 (1-10) Day #2 Homework: Page 396-397, 11-14, 27- 37 (odd)

What you’ll learn about… Area between curves Ares enclosed by intersecting curves Boundaries with changing functions Integrating with respect to y Saving time with geometric formulas.

Area Between Curves If f and g are continuous with f(x) > g(x) through [a,b], then the area between the curves y = f(x) and y = g(x) from a to b is the integral of [f-g] from a to b,

Applying the Definition Find the area of the region between the and y = sec 2 x and y = sinx from x = 0 to x = π/4. When graphed, you can see that y = sec 2 x is above y = sinx on [0, π/4].

Area of an Enclosed Region When a region is enclosed by intersecting curves, the intersection points give the limits of integration. Find the area of the region enclosed by the parabola y = 2 – x 2 and the line y = -x. Solution: graph the curves to determine the x-values of the intersections and if the parabola or the line is on top.

Using a Calculator Find the area of the region enclosed by the graphs of y = 2cosx and y = x 2 -1. Solution: Graph to determine the intersections. Don’t forget to use Zoom 7: Trig (radian mode, please)  X = ± 1.265423706  fnInt(2cosx –(x 2 -1), x, -1.265423706, 1.265423706)  4.994 units 2

Boundaries with Changing Functions If a boundary of a region is defined by more than one function, we can partition the region into subregions that correspond to the function changes. Find the area of the region R in the first quadrant bounded by y = (x) 1/2 and below by the x-axis AND the line y = x -2 Region A: Region B:

Drill: Find the x and y coordinates of all points where the graphs of the given functions intersect y = x 2 – 4x and x + 6 y = e x and y = x + 1 y = x 2 – πx and y = sin x (-1, 5) and (6, 12) (0,1) (0,0)and (3.14, 0)

Integrating with Respect to y Sometimes the boundaries of a region are more easily described by functions of y than by functions of x. We can use approximating rectangles that are horizontal rather than vertical and the resulting basic formula has y in place of x.

Integrating with Respect to y Find the area of the region R in the first quadrant bounded by y = (x) 1/2 and below by the x-axis AND the line y = x -2 by integrating with respect to x. y = x – 2 becomes y + 2 = x and y = (x) 1/2 becomes y 2 = x, y > 0

Making the Choice Find the area of the region enclosed by the graph y = x 3 and x = y 2 – 2 (To graph…. It makes more sense to integrate with respect to y because you will not have to split the region. The lower limit is when the y – value = -1 The upper limit needs to be found in the calc. y = 1.79

Solution Solve both equations for x.  y 1/3 = x and x = y 2 -2 Set up your definite integral: right – left Evaluate using the calculator.  fnInt(x 1/3 – x 2 + 2, x, -1, 1.79)  4.21 units 2

Using Geometry Another way to complete find the area of the region R in the first quadrant bounded by y = (x) 1/2 and below by the x-axis AND the line y = x -2 by integrating with respect to x. You can find the area under y = x to the x-axis from 0 to 4 and then subtract the area of the triangle formed by the line y = x -2 and the x-axis.

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