Download original notes
PAGE 1

5.4 Working with Integrals

We know \[ \int_{a}^{b} f(x) dx \] gives us the area between \( f(x) \) and x-axis on \( [a, b] \).

A small coordinate graph showing a curve above and below the x-axis with plus and minus signs.
  • if \( f(x) \ge 0 \), then area is positive
  • if \( f(x) \le 0 \), then area is negative

Even Functions

if \( f(x) \) is even: \( f(-x) = f(x) \rightarrow \) y-axis symmetry

Graph of an even function symmetric about the y-axis with shaded areas from -a to 0 and 0 to a.

notice

\[ \int_{-a}^{0} f(x) dx = \int_{0}^{a} f(x) dx \]

and since

\[ \int_{-a}^{a} f(x) dx = \int_{-a}^{0} f(x) dx + \int_{0}^{a} f(x) dx \]

so,

\[ \int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx = 2 \int_{-a}^{0} f(x) dx \]

so, if \( f(x) \) is even, find area of half then multiply by 2

PAGE 2

Odd Functions

if \( f(x) \) is odd: \( f(-x) = -f(x) \rightarrow \) origin symmetry

Graph of an odd function with origin symmetry, showing equal shaded areas above and below the x-axis.

notice

\[ \int_{-a}^{0} f(x) dx = -\int_{0}^{a} f(x) dx \]

so,

\[ \int_{-a}^{a} f(x) dx = \int_{-a}^{0} f(x) dx + \int_{0}^{a} f(x) dx \] \[ = 0 \]

so, for odd function, the net area from \( x = -a \) to \( x = a \) is 0.

PAGE 3

Example: Definite Integral of an Even Function

Calculate the definite integral:

\[ \int_{-1}^{1} (x^4 + 3) \, dx \]

This represents the area bounded by \( f(x) = x^4 + 3 \) and the x-axis from \( x = -1 \) to \( x = 1 \).

Symmetry Analysis

Notice \( f(x) = x^4 + 3 \) is even because:

\[ f(-x) = (-x)^4 + 3 = x^4 + 3 = f(x) \]

So we know:

\[ \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx \]

Calculation

Instead of calculating \( \int_{-1}^{1} (x^4 + 3) \, dx \), we can do:

\[ 2 \int_{0}^{1} (x^4 + 3) \, dx \quad \text{or} \quad 2 \int_{-1}^{0} (x^4 + 3) \, dx \]
\[ = 2 \left. \left( \frac{x^5}{5} + 3x \right) \right|_{0}^{1} = 2 \left( \frac{1}{5} + 3 \right) - 2(0) = \frac{32}{5} \]
Final Answer: \( \frac{32}{5} \)
PAGE 4

Example: Definite Integral of an Odd Function

Calculate the definite integral:

\[ \int_{-\pi}^{\pi} \sin x \, dx \]
Sketch of the sine function from -pi to pi, showing equal areas above and below the x-axis.

Symmetry Analysis

Since \( \sin x \) is odd, we know:

\[ \int_{-a}^{0} \sin x \, dx = -\int_{0}^{a} \sin x \, dx \]
\[ \int_{-\pi}^{\pi} \sin x \, dx = \int_{-\pi}^{0} \sin x \, dx + \int_{0}^{\pi} \sin x \, dx = 0 \]

Check

\[ \int_{-\pi}^{\pi} \sin x \, dx = -\cos x \Big|_{-\pi}^{\pi} \]\[ = (-\cos \pi) - (-\cos(-\pi)) \]\[ = (-(-1)) - (-(-1)) = 1 - 1 = 0 \]
PAGE 5

Average value of \( f(x) \) on \( [a, b] \)

What is the average value of \( f(x) \)?

A graph of a function f(x) on the interval [a, b] with axes labeled x and y.

Idea: replace \( f(x) \) with a constant function such that the area underneath is the same as area under \( f(x) \)

Graph showing the shaded area under a curve f(x) from a to b.
=
Graph showing a rectangle with height f_av and width (b-a) with the same area.
\[ \int_{a}^{b} f(x) dx = f_{av} (b - a) \]
\[ f_{av} = \frac{1}{b-a} \int_{a}^{b} f(x) dx \]
PAGE 6

Example: \( f(x) = \cos x \) on \( [0, \frac{\pi}{2}] \)

Graph of cos(x) from 0 to pi/2, with a horizontal line indicating the average value f_av.
\[ \begin{aligned} f_{av} &= \frac{1}{b-a} \int_{a}^{b} \cos x \, dx \\ &= \frac{1}{\frac{\pi}{2} - 0} \int_{0}^{\frac{\pi}{2}} \cos x \, dx \\ &= \frac{1}{\frac{\pi}{2}} (\sin x) \Big|_0^{\frac{\pi}{2}} \\ &= \frac{2}{\pi} (\sin \frac{\pi}{2} - \sin 0) \\ &= \frac{2}{\pi} \end{aligned} \]

Note: average is not necessarily the middle between max and min.

PAGE 7

5.5 Substitution Rule (part 1)

Integration by Substitution is the reverse of Chain Rule.

\[ f(x) = \frac{1}{4}(x^2 + 3)^4 \quad \text{when differentiating, let } (x^2 + 3) = u \]\[ = \frac{1}{4}u^4 \quad u = x^2 + 3 \]\[ f'(x) = \frac{1}{4} \cdot 4u^3 \cdot \frac{du}{dx} = u^3 \cdot 2x = (x^2 + 3)^3 \cdot 2x \]

When we want to find the antiderivative of \( (x^2 + 3)^3 \cdot 2x \) we need to undo the Chain Rule.

\[ \int (x^2 + 3)^3 (2x) dx \]

let \( u = x^2 + 3 \) just like how we got here

then \( \frac{du}{dx} = 2x \) and multiplying by \( dx \) we get

\[ du = 2x \, dx \]
"differential of u"
PAGE 8
\[ \int \underbrace{(x^2 + 3)^3}_{u^3} \underbrace{(2x) dx}_{du} \]
\[ = \int u^3 du \quad \text{handle this just like } \int x^3 dx \]\[ = \frac{u^4}{4} + C \quad \text{now undo the substitution } u = x^2 + 3 \]
\[ = \frac{1}{4}(x^2 + 3)^4 + C \]

the main challenge of substitution is finding \( u \)

in this example we had Chain Rule to look at to know what \( u \) is, but generally we need to find it

PAGE 9

How do we choose \( u \)?

Pick the part of the integrand (the things between \( \int \) and \( dx \)) such that its derivative is a constant multiple of the other part.

Previous Example:

\[ \int (x^2 + 3)^3 (2x) \, dx \]

Derivative of \( x^2 + 3 \) is a constant (1) multiple of the other part \( (2x) \).

Easier Rule of Thumb:

Choose the more complicated part (or part with higher power) to be \( u \).

PAGE 10

Example

\[ \int x^2 \sqrt{x^3 + 4} \, dx \]

\[ = \int (x^2) (x^3 + 4)^{1/2} \, dx \]

Compare \( x^2 \) and \( x^3 + 4 \).

\( x^3 + 4 \) is more complicated, so let's pick it to be \( u \).

\[ u = x^3 + 4 \]

Then take its derivative: \( \frac{du}{dx} = 3x^2 \)

Then find the differential of \( u \):

\[ du = 3x^2 \, dx \]

Now remove all \( x \) and \( dx \) from original integral:

\[ \int (x^3 + 4)^{1/2} \]\[ \downarrow \]\[ u^{1/2} \]
\[ (x^2) \, dx \]\[ \downarrow \]\[ \frac{1}{3} du \]

Substitution Logic:

\[ du = 3x^2 \, dx \]

\[ \frac{1}{3} du = x^2 \, dx \]

PAGE 11
\[ = \int u^{1/2} \cdot \frac{1}{3} du = \frac{1}{3} \int u^{1/2} du \]

treat it like \( \int x^{1/2} dx \)

\[ = \frac{1}{3} \left( \frac{u^{3/2}}{3/2} \right) + C = \frac{1}{3} \cdot \frac{2}{3} u^{3/2} + C \]

\( u = x^3 + 4 \)

undo the subs

\[ = \frac{2}{9} (x^3 + 4)^{3/2} + C \]

PAGE 12

example

\[ \int e^{2x+3} dx \]
\[ = \int e^{2x+3} (1) dx \]

compare to power of \( e \) to the other part

\( 2x+3 \) vs. \( 1 \)

here, let \( u = 2x+3 \) because it's more complicated.

\( \frac{du}{dx} = 2 \)

\( du = 2 dx \)
\[ \int e^{2x+3} dx \]

\( \downarrow \)
\( e^u \)

\( \downarrow \)
\( \frac{1}{2} du \)

we only have \( dx \) so divide by 2

\( \frac{1}{2} du = dx \)

\[ = \int \frac{1}{2} e^u du = \frac{1}{2} \int e^u du = \frac{1}{2} e^u + C \]

\[ = \frac{1}{2} e^{2x+3} + C \]

PAGE 13

Definite integrals: change back to x then evaluate

Example

\[ \int_{0}^{1} (x^2 + 3)^3 (2x) dx \]

pick u as usual:

\[ u = x^2 + 3 \]\[ \frac{du}{dx} = 2x \]\[ du = 2x dx \]
\[ \int_{x=0}^{x=1} u^3 du \]

Integration limits are for x NOT u

\[ = \left. \frac{1}{4} u^4 \right|_{x=0}^{x=1} \]

change back to x by removing u

\[ = \left. \frac{1}{4} (x^2 + 3)^4 \right|_{x=0}^{x=1} = \frac{1}{4} (4)^4 - \frac{1}{4} (3)^4 = \frac{175}{4} \]
Result: \( \frac{175}{4} \)
PAGE 14

Other way to do definite integrals: change limits from x to u

\[ \int_{0}^{1} (x^2 + 3)^3 (2x) dx \]
\[ u = x^2 + 3 \]\[ du = 2x dx \]
\[ \int_{x=0}^{x=1} \underbrace{(x^2 + 3)^3}_{u^3} \underbrace{(2x) dx}_{du} \]

Now we adjust limits so the numbers refer to u

From \( u = x^2 + 3 \)

old upper limit: \( x = 1 \rightarrow u = (1)^2 + 3 = 4 \)

old lower limit: \( x = 0 \rightarrow u = (0)^2 + 3 = 3 \)

\[ = \int_{u=3}^{u=4} u^3 du = \left. \frac{u^4}{4} \right|_{3}^{4} = \frac{4^4}{4} - \frac{3^4}{4} = \frac{175}{4} \]