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)
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
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)
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
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
Integration by Parts udv = uv vdu Template: Choose: u = part which gets simpler after differentiation dv = easy to integrate part Compute: du = v =
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
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
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
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 1 0.5 f (x) = e x 2-2.4-2 -1.6-1.2-0.8-0.4 0 0.4 0.8 1.2 1.6 2 2.4 T 2 (x) =1 x 2
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 0 1 0 for x-values near 0
Integration Using Taylor Polynomials We can obtain a better approximation by using a higher degree Taylor polynomial to represent the integrand. 1.25 1 0.75 0.5 Example: 1 0 e x2 dx 1 0 (1 x 2 + 1 2 x 4 1 6 x6 ) dx x 1 3 x3 + 1 10 x5 1 42 x 7 0.74286 1 0 0.25-2.4-2 -1.6-1.2-0.8-0.4 0 0.4 0.8 1.2 1.6 2 2.4
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)
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
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
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
Application Example: Several very skinny 2.0-m-long snakes are collected in the Amazon. Each snake has a density of ρ(x) =1+ 2 10 8 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
Application ρ(x) ρ(x) 5 1 2.5 0.75-500 -250 0 250 500 750 x 0.5-2.5 0.25 0 25 50 75 100 125 150 175 x200 x
Application (a) Find the total mass of each snake. (b) Find the average density of each snake.
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.
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
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