1. Lesson 4-10b Anti-Differentiation

2. Quiz Estimate the area under the graph of f(x) = x² + 1 from x = -1 to x = 2 …. Improve your estimate by using six right endpoint rectangles.

3. Objectives Understand the concept of an antiderivative Understand the geometry of the antiderivative and that of slope fields Work rectilinear motion problems with antiderivatives

4. Vocabulary Antiderivative – the opposite of the derivative, if f(x) = F’(x) then F(x) is the antiderivative of f(x) Integrand – what is being taken the integral of [F’(x)] Variable of integration – what variable we are taking the integral with respect to Constant of integration – a constant (derivative of which would be zero) that represents the family of functions that could have the same derivative

5. Two other Anti-derivative Forms 1.Form: 2.Form: F(x) = e x dx = e x + C ∫ 1 F(x) = ----- dx = ln |x| + C x ∫ Remember derivative of e x is just e x Remember derivative of ln x is (1/x)

6. Practice Problems a) (2e x + 1) dx ∫ b) (1 – x -1 ) dx ∫ -5 c) ----- dx x ∫ d) (e x + x² - 1) dx ∫ 2e x + x + C e x + ⅓x 3 - x + C x – ln |x| + C -5 ln|x| + C

7. How to Find C In order to find the specific value of the constant of integration, we need to have an initial condition to evaluate the function at (to solve for C)! Example: find such that F(1) = 4. F(x) = (2x + 3) dx ∫ F(x) = x² + 3x + C F(1) = 4 = (1)² + 3(1) + C = 4 + C 0 = C (boring answer!)

8. Acceleration, Velocity, Position Remember the following equation from Physics: s(t) = s 0 + v 0 t – ½ at² where s 0 is the initial offset distance (when t=0) v 0 is the initial velocity (when t=0) and a is the acceleration constant (due to gravity) We can solve problems given either s(t) or a(t) (and some initial conditions) – basically solving the problem from either direction!

9. Motion Problems Find the velocity function v(t) and position function s(t) corresponding to the acceleration function a(t) = 4t + 4 given v(0) = 8 and s(0) = 12. v(t) = (4t + 4) dt = 2t² + 4t + v 0 ∫ v(0) = 8 = 2(0)² + 4(0) + v 0 8 = v 0 = 2t² + 4t + 8 s(t) = (2t² + 4t + 8) dt = ⅔t³ + 2t² + 8t + s 0 ∫ s(0) = 12 = ⅔(0)³ + 2(0)² + 8(0) + s 0 12 = s 0 s(t) = ⅔t³ + 2t² + 8t + 12

10. Motion Problems A ball is dropped from a window hits the ground in 5 seconds. How high is the window (in feet)? v(t) = a(t) dt = -32t + v 0 ∫ v(0) = 0 = -32(0) + v 0 (ball was dropped) 0 = v 0 s(t) = v(t) dt = -16t² + s 0 ∫ s(5) = 0 = -16(5)² + s 0 s 0 = 400 feet window was 400 feet up a(t) = - 32 ft/s²

11. Slope Fields A slope field is the slope of the tangent to F(x) (f(x) in an anti-differentiation problem) plotted at each value of x and y in field. Since the constant of integration is unknown, we get a family of curves. An initial condition allows us to plot the function F(x) based on the slope field.

12. f(0) = -2 Slope Field Example

13. Summary & Homework Summary: –Anti-differentiation is the reverse of the derivative –It introduces the integral –One of its main applications is area under the curve Homework: –pg 358-360: Day 1: 1-3, 12, 13, 16 Day 2: 25, 26, 53, 61, 74