Drill Find the area between the xaxis and
Drill • Find the area between the x-axis and the graph of the function over the given interval: • y = sinx over [0, π] • y = 4 x-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, 2737 (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 = 2 cosx 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 § fn. Int(2 cosx –(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: A B A +B = 3. 33
Drill: Find the x and y coordinates of all points where the graphs of the given functions intersect • y = x 2 – 4 x and x + 6 • y = ex 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 x = y 2 – 2 (To graph…. y = x 3 and • 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. § fn. Int(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. 2 10/3 units
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