Estimate Area Under Graph using Right Endpoint Rectangles in Anti-Differentiation

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 F(x) = ----- dx = ln |x| + C x F(x) = ex dx = ex + C • Form: • Form: Remember derivative of ex is just ex Remember derivative of ln x is (1/x)

6. Practice Problems ∫ ∫ ∫ ∫ -5 c) ----- dx x a) (2ex + 1) dx b) (1 – x-1) dx d) (ex + x² - 1) dx 2ex + x + C -5 ln|x| + C ex + ⅓x3 - x + C x – 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) = s0 + v0t – ½ at² where s0 is the initial offset distance (when t=0)v0 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 + v0 = 2t² + 4t + 8 v(0) = 8 = 2(0)² + 4(0) + v0 8 = v0 ∫ s(t) = (2t² + 4t + 8) dt = ⅔t³ + 2t² + 8t + s0 s(0) = 12 = ⅔(0)³ + 2(0)² + 8(0) + s0 12 = s0 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)? a(t) = - 32 ft/s² ∫ v(t) = a(t) dt = -32t + v0 v(0) = 0 = -32(0) + v0 (ball was dropped) 0 = v0 ∫ s(t) = v(t) dt = -16t² + s0 s(5) = 0 = -16(5)² + s0 s0 = 400 feet window was 400 feet up

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. y Slope Field Example f(0) = -2 x

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