Announcements. Topics: Homework:

Transcription

1 Announcements Topics: - sections 7.3 (the definite integral +area), 7.4 (FTC), 7.5 (additional techniques 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 Types of Integrals Indefinite Integral function of x f (x) dx = F(x) + C antiderivative of f Definite Integral b a f (x) dx = net area number

3 The Fundamental Theorem of Calculus If f is continuous on [a, b], then b a f (x) dx = F(x) a b = F(b) F(a) where F is any antiderivative of f, i.e., F'= f.

4 Evaluating Definite Integrals Example: Evaluate each definite integral using the FTC. (a) 3 0 (x ) dx (b) 2 ( 4 t + t 4 )dt 2 4 4x dx 2 (c) (d) 2 (3x ) 2 x dx

5 Evaluating Definite Integrals Example: Try to evaluate the following definite integral using the FTC. What is the problem? 4 (x 2) 2 dx

6 Differentiation and Integration as Inverse Processes If f is integrated and then differentiated, we arrive back at the original function f. d dx x a f (t) dt = f (x) FTC I If F is differentiated and then integrated, we arrive back at the original function F. b a d dx F(x) dx = F(x) b a FTC II

7 The Definite Integral - Total Change Interpretation: The definite integral represents the total amount of change during some period of time. Total change in F between times a and b: value at end F(b) F(a) = value at start b a df dt dt rate of change

8 Application Total Change Example: Suppose that the growth rate of a fish is given by the differential equation dl dt = 6.48e 0.09t where t is measured in years and L is measured in centimetres and the fish was 0.0 cm at age t=0 (time measured from fertilization).

9 Application Total Change (a) Determine the amount the fish grows between 2 and 5 years of age. (b) At approximately what age will the fish reach 45cm?

10 The Chain Rule and Integration by Recall: Substitution The chain rule for derivatives allows us to differentiate a composition of functions: derivative [ f (g(x))]' = f '(g(x))g'(x) antiderivative

11 The Chain Rule and Integration by Substitution Suppose we have an integral of the form where f (g(x))g'(x)dx F'= f. composition of functions derivative of Inside function F is an antiderivative of f Then, by reversing the chain rule for derivatives, we have f (g(x))g'(x)dx = F(g(x)) + C. integrand is the result of differentiating a composition of functions

12 Example Integrate (x 3 +) 4 3x 2 dx.

13 Integration by Substitution Algorithm:. Let u = g(x) where g(x) is the part causing problems and g'(x) cancels the remaining x terms in the integrand. 2. Substitute u = g(x) and du = g'(x)dx into the integral to obtain an equivalent (easier!) integral all in terms of u. f (g(x))g'(x)dx = f (u)du

14 Integration by Substitution Algorithm: 3. Integrate with respect to u, if possible. f (u)du = F(u) + C 4. Write final answer in terms of x again. F(u) + C = F(g(x)) + C

15 Integration by Substitution Example: Integrate each using substitution. (a) 2x + 5 x 2 + 5x 7 dx (b) xe 4 x 2 dx cos x (c) (d) x dx e 4 ln x x dx

16 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

17 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)

18 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

19 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

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

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

22 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

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

24 Approximating Functions with Polynomials Recall: The quadratic approximation to f (x) = e x 2 around the base point x=0 is T 2 (x) = x 2. base point 0.5 f (x) = e x T 2 (x) = x 2

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

26 Integration Using Taylor Polynomials We can obtain a better approximation by using a higher degree Taylor polynomial to represent the integrand Example: 0 e x2 dx 0 ( x x 4 6 x6 ) dx x 3 x3 + 0 x5 42 x