advertisement

Drill • Find the area between the x-axis and the graph of the function over the given interval: • y = sinx over [0, π] sin xdx 0 cos x 0 cos cos 0 2 • y= 4x-x3 over [0,3] 2 3 (4 x x ) dx 3 0 2 2 2x 2 x 4 4 0 (4 x x 3 2x 2 2 4 6 . 25 10 . 25 x 4 4 3 ) dx 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, b A [ f ( x ) g ( x )] dx a Applying the Definition • Find the area of the region between the and y = sec2 x and y = sinx from x = 0 to x = π/4. • When graphed, you can see that y = sec2 x is above y = sinx on [0, π/4]. /4 (sec 2 /4 x sin x ) dx 0 tan x cos x 0 (tan 4 cos ) 1 4 1 2 (tan 0 cos 0 ) 0 1 1 2 2 2 1 2 2 units 2 • • • 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 – x2 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. 2 [( 2 x ) ( x )] dx 2 1 2 ( 2 x x ) dx 2 1 2 2x 1 x 3 3 x 2 2 ( 4 8 / 3 2 ) ( 2 1 / 3 1 / 2 ) 5 . 5 units 2 Using a Calculator • Find the area of the region enclosed by the graphs of y = 2cosx and y = x2-1. • Solution: Graph to determine the intersections. Don’t forget to use Zoom 7: Trig (radian mode, please) X = ± 1.265423706 fnInt(2cosx –(x2-1), x, -1.265423706, 1.265423706) 4.994 units2 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 2 Region A: 2 2 (x 1/ 2 ) dx 0 x 1 . 8856 3/2 0 3 4 Region B: (x 1/ 2 ( x 2 )) dx 2 4 2 2 3 x 3/2 1 x 2x 2 2 1 . 4447733 Drill: Find the x and y coordinates of all points where the graphs of the given functions intersect • y = x2 – 4x and x + 6 • y = ex and y = x + 1 • y = x2 – 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 y2 = x, y > 0 2 ( y 2 y ) dy 2 0 2 y 0 (2) 2 2 2y 2 2 2(2) y 3 3 (2) 3 3 3 (0) 2 (0) 2(0) 3 2 Making the Choice • Find the area of the region enclosed by the graph 3 x = y2 – 2 (To graph…. y 1 x y2 x2 y = x3 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. y1 x y2 y = 1.79 3 x2 Solution • Solve both equations for x. y1/3 = x and x = y2 -2 • Set up your definite integral: right – left 1 . 79 (y 1/3 ( y 2 )) dy 2 0 • Evaluate using the calculator. fnInt(x1/3 – x2 + 2, x, -1, 1.79) 4.21 units2 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. 4 4 (x 4 3 0 x 3/2 x 2 ) dx 0 4 2 1/ 2 2 2 2 2x ( x 2 ) dx ( 2 3 4 3/2 2 42 2 ) ( 2 4) ( 2 2) 2 2