ch 5.6: method of substituion; newton’s law of...

Ch 5.6: Method of Substituion; Newton’s Law of Cooling

Theorem

∫f (g(x)) · g ′(x) dx =

∫f (u) du

where f is a continuous function on the range of u = g(x).

Example) Find∫

(x + 1)4 dx .

ExamplesEvaluate the following integrals.

1.∫ √

1 + y 2 · 2y dy

2.∫

2x(1 + x2)10

dx

Integral of the tangent and secant functionI∫

tan x dx = − ln | cos x |+ C = ln | sec x |+ CI∫

sec x dx = ln | sec x + tan x |+ C

Example

Find∫

1cos2(3x)

dx .

u?Find

∫csc2(2θ) cot(2θ) dθ by letting

1. u = cot(2θ)2. u = csc(2θ).

Using an identity

Find∫

cos2(x) dx .

The Substitution Rule for Definite integrals

TheoremIf g

′(x) is continuous on [a, b] and f is a continuous function on

the range of u = g(x), then∫ b

af (g(x)) · g ′(x) dx =

∫ g(b)

g(a)f (u) du.

Example) Evaluate∫ 4

0

√2x + 1 dx

Example

Evaluate∫ 2

11

(3−5x)2 dx

Integrals of Even and Odd functionsLet f be continuous on [−a, a], a > 0.

I If f is an even function, then∫ a−a f (x) dx = 2

∫ a0 f (x) dx

I If f is an odd function, then∫ a−a f (x) dx = 0

ExamplesEvaluate the following integrals.

1.∫ 2−2 (3− x2) dx

2.∫ 2−2 x(3− x2) dx

Newton’s Law of Cooling (Also taught in Math 2306)Let u(t) denote the temperature of an object at time t, if T is theconstant room temperature, then

du

dt= k[u(t)− T ]

Its solution is given by

u(t) = (u0 − T )ekt + T

where u0 = u(t = 0), and k is constant.

ExampleWhen a cake is just removed from an oven, its temperature ismeasured at 300◦ F. Three minutes later its temperature is 200◦ F.How long will it take for the cake to cool off to 80◦ F? Assume theroom temperature is 70◦ F.

Class ExerciseEvaluate the following integrals. Show all steps.∫

1

x ln(x6)dx