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