Lesson 4-10b Anti-Differentiation. Quiz Estimate the area under the graph of f(x) = x² + 1 from x =...
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Transcript of Lesson 4-10b Anti-Differentiation. Quiz Estimate the area under the graph of f(x) = x² + 1 from x =...
Lesson 4-10b
Anti-Differentiation
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.
Objectives
• Understand the concept of an antiderivative
• Understand the geometry of the antiderivative and that of slope fields
• Work rectilinear motion problems with antiderivatives
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
Two other Anti-derivative Forms
1. Form:
2. Form:
F(x) = ex dx = ex + C ∫
1F(x) = ----- dx = ln |x| + C x∫
Remember derivative of ex is just ex
Remember derivative of ln x is (1/x)
Practice Problems
a) (2ex + 1) dx∫
b) (1 – x-1) dx∫
-5c) ----- dx x∫
d) (ex + x² - 1) dx∫
2ex + x + C
ex + ⅓x3 - x + Cx – ln |x| + C
-5 ln|x| + C
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!)
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!
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∫v(0) = 8 = 2(0)² + 4(0) + v0 8 = v0
= 2t² + 4t + 8
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
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 + 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
a(t) = - 32 ft/s²
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.
y
x
f(0) = -2
Slope Field Example
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