Lecture 5 §7.1 §7.8 Improper

Lecture 5 Integration Integration by parts

Recall the for differentiation d [f(x)g(x)] = f 0(x)g(x) + f(x)g0(x). dx Thus, Z f(x)g(x) = [f 0(x)g(x) + f(x)g0(x)] dx.

Formular for integration by parts: Z Z f(x)g0(x) dx = f(x)g(x) − f 0(x)g(x) dx

Let u = f(x) and v = g(x). The formula becomes Z Z udv = uv − vdu

Lecture 5 Integration Sometimes you need to apply the integration by parts formula more than once. Example(7.1-1) Compute R x sin x dx Exercise(7.1-5) Compute R xex/2 dx Example(7.1-3) Compute R t2et dt

Lecture 5 Integration Definite integrals

Thanks to the Fundamental Theorem of we can compute definite using the integration by parts formula

Z b Z b 0 b 0 f(x)g (x) dx = f(x)g(x)|a − f (x)g(x) dx a a

Lecture 5 Integration R 1 Example(7.1-5). Calculate 0 arctan x dx 0 1 (Hint: (arctan x) = 1+x2 ) R 1 2 −x Exercise(7.1-24) Compute 0 (x + 1)e dx

Lecture 5 Integration LIATE principle of integration by parts

A rule of thumb suggests to choose the function that comes first in the following list as u: Logarithmic: ln x Inverse trigonometric: arcsin x, arccos x, arctan x Algebraic: x2 Trigonometric: sin x, etc Exponential: ex

Lecture 5 Integration Some strategy

Simplify the integrand, for example R x(x + 1) dx = R (x2 + x) dx; R 2 R 1 R tan θ cos θ dθ = sin θ cos θ dθ = 2 sin 2θ dθ. Look for substitution R sin 2θ dθ, 2θ = u.

Try integration by parts.

Relate to previous exercises!

Lecture 5 Integration R q 1−x Example (7.5-5) Solve 1+x dx

R 3 4 Exercise (7.5-9) Evaluate 1 x ln x dx

Lecture 5 Integration Infinite intervals

To determine the area which lies under y = 1/x2, above the x-axis and to the right of x = 1, we have to compute an integral of this form Z ∞ 1 2 dx 1 x

For all t > 1 Z t 1 1 2 dx = 1 − < 1 1 x t

Lecture 5 Integration We have Z ∞ 1 Z t 1 2 dx = lim 2 dx = 1 1 x t→∞ 1 x

Lecture 5 Integration Improper integral of type 1: infinite intervals

R t If a f(x) dx exists for all t ≥ a, then Z ∞ Z t f(x) dx = lim f(x) dx a t→∞ a if this exists and is a finite number. R b The integral −∞ f(x) dx is defined in a similar way. R ∞ R b a f(x) dx and −∞ f(x) dx are called convergent if the limit exists and divergent if it does not. R ∞ R b If a f(x) dx and −∞ f(x) dx are convergent, we define Z ∞ Z t Z a f(x) dx = f(x) dx + f(x) dx. −∞ a −∞ Note: a is arbitrary.

Lecture 5 Integration R ∞ 1 Example (7.8-1) Determine 1 x dx Example (7.8-4) For which p is the following integral convergent?

Z ∞ 1 p dx 1 x

Lecture 5 Integration Improper integral of type 2: discountinuous integrands

We can define the improper integral also when a function f has a discontinuity (for example an asymptote)

Lecture 5 Integration If f is continuous in [a, b) and discontinuous at b Z b Z t f(x) dx = lim f(x) dx a t→b− a if this limit exists and is a finite number. If f is continuous in (a, b] and discontinuous at a Z b Z b f(x) dx = lim f(x) dx a t→a+ t if this limit exists and is a finite number. R b a f(x) dx is called convergent if the limit exists and divergent if it does not. R c If f has a discontinuity at c ∈ [a, b] and both a f(x) dx and R b c f(x) dx are convergent, we define Z b Z c Z a f(x) dx = f(x) dx + f(x) dx. a a c

Lecture 5 Integration 5 Example (7.8-5) Find R √ 1 dx 2 x−2 R 3 1 Example (7.8-5) Evaluate (if possible) 0 x−1 dx

Lecture 5 Integration Comparison Theorem

When you are not interested in the evaluating an integral but only in knowing whether it converges or not you can use the following

Suppose f and g are continuous functions with f(x) ≥ g(x) ≥ 0 for x ≥ a.

R ∞ R ∞  If 0 f(x) dx is convergent, then 0 g(x) dx is convergent

R ∞ R ∞  If 0 g(x) dx is divergent, then 0 f(x) dx is divergent

Lecture 5 Integration R ∞ 1 Exercise (7.8-25) Determine if 1 x(ln x)3 dx is convergent or divergent. If it converges evaluate it. R ∞ 1 Exercise (7.8-58) Find the values of p for which e x(ln x)p dx converges and evaluate the integral for those values of p.

Lecture 5 Integration