Announcements. Topics: Homework:

Transcription

1 Announcements Topics: - sections 7.5 (additional techniques of integration), 7.6 (applications of integration), * Read these sections and study solved examples in your textbook! Homework: - review lecture notes thoroughly - work on practice problems from the textbook and assignments from the coursepack as assigned on the course web page (under the SCHEDULE + HOMEWORK link)

2 The Product Rule and Integration by Parts The product rule for derivatives leads to a technique of integration that breaks a complicated integral into simpler parts. Integration by Parts Formula: udv = uv vdu given integral that we cannot solve hopefully this is a simpler Integral to evaluate

3 The Product Rule and Integration by Parts Deriving the Formula Start by writing out the Product Rule: d dx [u(x) v(x)] = du dx v(x) + u(x) dv dx Solve for u(x) dv dx : u(x) dv dx = d dx [u(x) v(x)] du dx v(x)

4 Deriving the Formula The Product Rule and Integration by Parts Integrate both sides with respect to x: u(x) dv dx dx = d dx [u(x) v(x)] dx v(x) du dx dx

5 Deriving the Formula The Product Rule and Integration by Parts Simplify: u(x) dv dx dx = d dx [u(x) v(x)] dx v(x) du dx dx u(x)dv = u(x) v(x) v(x) du

6 Integration by Parts udv = uv vdu Template: Choose: u = part which gets simpler after differentiation dv = easy to integrate part Compute: du = v =

7 Integration by Parts Example: Evaluate each using integration by parts. (a) x cos4 xdx (b) x 2 e x 2 dx (c) 2 1 ln x dx

8 Strategy for Integration Method Basic antiderivative Applies when the integrand is recognized as the reversal of a differentiation formula, such as Guess-and-check the integrand differs from a basic antiderivative in that x is replaced by ax+b, for example Substitution both a function and its derivative (up to a constant) appear in the integrand, such as Integration by parts the integrand is the product of a power of x and one of sin x, cos x, and e x, such as the integrand contains a single function whose derivative we know, such as

9 Strategy for Integration What if the integrand does not have a formula for its antiderivative? Example: impossible to integrate 1 0 e x 2 dx

10 Approximating Functions with Polynomials Recall: The quadratic approximation to f (x) = e x 2 around the base point x=0 is T 2 (x) =1 x 2. base point f (x) = e x T 2 (x) =1 x 2

11 Integration Using Taylor Polynomials We approximate the function with an appropriate Taylor polynomial and then integrate this Taylor polynomial instead! Example: impossible to integrate 1 easy to integrate e x 2 dx (1 x 2 ) dx for x-values near 0

12 Integration Using Taylor Polynomials We can obtain a better approximation by using a higher degree Taylor polynomial to represent the integrand Example: 1 0 e x2 dx 1 0 (1 x x x6 ) dx x 1 3 x x x

13 The Definite Integral Area Between Curves The area between the curves y = f (x) and between and is y = g(x) x = a x = b and A = b a f (x) g(x) dx Recall: f (x) g(x) = f (x) g(x) when f (x) g(x) g(x) f (x) when f (x) g(x)

14 The Definite Integral Area Between Curves Examples: Sketch the region enclosed by the given curves and then find the area of the region. (a) y = x 2 2x, y = x + 4 (b) y = x, y = 1 x, x = 1 2, x = 2

15 The Definite Integral - Average Value The average value of a function f on the interval from a to b is f (x) f = 1 b a b a f (x) dx f For a positive function, average height = area width

16 The Definite Integral - Average Value f (x) area = b a f (x)dx area = base height = (b a) f f b a f (x) dx = (b a) f

17 Application Example: Several very skinny 2.0-m-long snakes are collected in the Amazon. Each snake has a density of ρ(x) = x 2 (300 x) where ρ is measured in grams per centimeter and is measured in centimeters from the tip of the tail. Find the average density of the snake. x

18 Application ρ(x) ρ(x) x x200 x

19 Application (a) Find the total mass of each snake. (b) Find the average density of each snake.

20 Approximating Volumes A(x i ) = area of base Δx = b a n V n = A(x 1 )Δx + A(x 2 )Δx +! + A(x n )Δx n i=1 = A(x i )Δx Riemann Sum So, the volume V of the solid S V n.

21 Integrals and Volumes Definition: Denote by A(x) the area of the cross-section of S by the plane perpendicular to the x-axis that passes through x. Assume that A(x) is continuous on [a,b]. Then the volume V of S is given by V = limv n = lim n n provided that the limit exists. n A(x i )Δx = A(x) dx i=1 b a

22 Volumes of Solids of Revolution Examples: Find the volume of the solid obtained by rotating the region R enclosed (bounded) by the given curves about the given axis. (a) y = 1 x, y = 0, x =1, and x = 2 about the x - axis (b) y = 8 x, y = 3, x = 2, and x = 5 about the y - axis