# Section 6.1 Volumes from Integration

Integrals can be used any time we accumulate something. It could be accumulating speed to get distance, lengths to get area, even area to get volume or force to get work.

Consider a rectangular box, $$l\cdot w \cdot h$$. Notice that the $$l\cdot w$$ is the area of the base. So, Volume = Area of Base x height. Consider the integral:

$$! \int_0^h (l\cdot w) \, dx =l\cdot w \cdot h$$

This works for any area as long as the the top and bottom are parallel and the cross-sections have the same area.

$$! \int_0^h \textrm{Area}\,dx = Volume$$

#### Example 1

Consider the functions $$y=\sqrt{x}$$ and $$y=-\sqrt{x}$$. Imaging a square, perpendicular to the $$x-y$$-plane and $$x$$-axis whose diagonals span between these curves.

The length of any diagonal would then be $$2 \sqrt{x}$$ and a side length would be $$\sqrt{2x}$$ Now let that square slide from $$x=0$$ to $$x=4$$. Let’s sketch this solid and find it’s volume.

Volume = $$\displaystyle \int_0^4 \textrm{side}^2\,dx = \int_0^4 \big( \sqrt{2x}\big)^2\,dx = \int_0^4 2x\,dx = \big[x^2\big]_0^4 = 16$$ square units.

### Rotational Volumes

We can also find volumes by rotating areas or accumulating discs

Consider rotating the line $$y=3$$ about the $$x$$-axis.

To get a cross-sectional area we consider that a circular disc has an area of $$\pi r^2$$ and here the radius is the distance from the $$x$$-axis to the line $$y=3$$.

$$! \int_{x=0}^5 \pi (3)^2\,dx = \pi\cdot 3^2 \cdot 5$$

We can rotate any curve about an axis to generate volumes of all sorts.

# Intersecting Cylinders with equal radii

Imaginw two pipes of the same size intersecting. What does that intersection look like?

It is slightly larger than a sphere of the same radius.

Since it is symmetric top to bottom, just consider the top half. Slices of this intersection, parallel to the plane the two pipes sit on are squares.

The length of each square is $$2\sqrt{r^2-y^2}$$ where $$y$$ is measured perpendicular to the plane of the pipes, starting at their common center.

A particular slice, parallel to the plane of the pipes, above the common center would then have an area of $$\bigg(2\sqrt{r^2-y^2}\bigg)^2$$.

If we integrate this area as $$y$$ goes from $$0$$ to $$r$$, then we will get the volume of the top half:

$$!\int_{y=0}^r \bigg(2\sqrt{r^2-y^2}\bigg)^2\,dy$$

# Relatively Prime Integers

a and b are said relatively prime if GCD(a,b)=1.

Euler’s Totient function, \phi(n), counts the number of positive integers less than n that are are relatively prime to n.

For example, numbers relatively prime to 15 are {1,2,4,7,8,11,13,14}, so \phi(15)=8.

1,  2,  3,  4,  5,  6,  7 8,  9, 10, 11, 12, 13, 14

There is always an interesting symmetry to these lists, that is, if a is in the list, then n-a is in the list.

Take a<n and GCD(a,n)=1.

Now lets suppose that n and n-a are not relatively prime, so GCD(n-a,n)=d>1.

This means that d|n so there exists a k where n=dk. Similarly, n-a=dj for some j.

Then a=n-dj=dk-dj=d(k-j), so that d|a, contradicting our hypotheses, GCD(a,n)=1.

So d must me 1.

This may also be shown using the fact that a and n are relatively prime, if and only if there exist integers x and y such that ax+ny=1.(ref)

So far, I lack an simpler argument that if a is relatively prime to n and a<n then n-a is also.

# Some Number Theory Basics

#### A linear combination theorem about divisors

If c|a and c|b, then c|(ax+by) for arbitrary integers x and y.

If c|a then there exists an integer n such that a=cn, similarly, b=cm for some m.

Given any integers x and y, ax+by = cnx+cmy = c(nx+my). Since n, m, x, and y are integers, then so is nx+my.

#### Concerning the gcd of a and b

Given integers a and b, there exist integers u and v such that au+bv= gcd(a,b).

Now consider the set S={ ax+by | ax+by > 0 and x and y are integers }

S is non empty since if y=0, then ax+b*0=|a| if we choose x to be 1 or -1 as necessary. The set S must contain some smallest number, say d. There exist u and v such that d=au+bv.

Now let’s apply the Division Algorithm. We can find q and r so that a=qd+r, where 0 <= r <= d.

So, r = a-qd = a-q(au+bv) = a(1-qu)+b(-qv).

Suppose r is positive. Then r is in S. This contradicts the Division Algorithm, since r<d. Hence r=0 and a=qd. Now, d|a and similarly, d|b, so d is a common divisor of a and b.

Take c to be any common divisor of a and b. c|(au+bv) and so c|d. This implies that c <= d and d must be the gcd(a,b).

Now that we know values exist, the Euclidean Algorithm for finding the gcd of two numbers can be used to find values for u and v.

#### Linear Combinations of Relatively Prime Integers

a and b are relatively prime, if and only if there exist integers x and y such that ax+by=1.

# Fractions

#### What is a fraction?

• part of a whole group
• part of a whole area
• position on a number line
• division
• ratio

#### Proper Fractions

numerator < denominator

#### Improper Fractions

numerator ≥ denominator

#### Fundamental Law of Fractions

a/b=(a*n)/(b*n)

#### Simplest form

GCD(a,b) = 1, relatively prime

#### Ordering and Comparison

a/b > c/d if ad > bc

#### Denseness

Between any two fractions, there are more fractions

# Section 5.6 Substitution and Area Between Curves

Substitution with integrals, that is, reversing the chain rule for derivatives, can help us find antiderivatives for special function combinations.

According to the fundamental theorem of calculus, we can also use those antiderivatives to find areas between curves.

#### Example 1

I want to find the area between $$f(x)=x^2-1$$ and $$g(x)=1-x^2$$ on the interval $$[-1,1]$$

Since $$g(x)$$ is the upper curve here, we want the integral $$\displaystyle \int_{-1}^1 \big(g(x)-f(x)\big)\,dx$$

Note this is the same as $$\displaystyle 2\cdot\int_{0}^1 \big(g(x)-f(x)\big)\,dx$$ since the functions have even symmetry.

Examples

• The area between $$y=1$$ and $$y=\cos^2(x)$$ on the interval $$0,\pi$$
• The area between $$y=0$$, $$x+y=2$$, and $$y=x^2$$ on the interval $$0,2$$

# Symmetry with Sums of Proper Fractions

If a is relatively prime to b, then b-a is also relatively prime to b. (ref)

Consider the sum a/b+c/d, where both fractions are reduced, that is, a is relatively prime to b and c is relatively prime to d.

Then the sum (b-a)/b+(d-c)/d is also the sum of two reduced fractions.

(b-a)/b+(d-c)/d=1-a/b+1-c/d=2-(a/b+c/d)

So if a/b+c/d is a reducible sum then so is (b-a)/b+(d-c)/d. Better yet, their sum is always 2.

# A substitution example

Consider $$\displaystyle \int{\sqrt{\frac{x-1}{x^5}}}\,dx$$

Without doing a bit of algebra on the integrand, most substitutions don’t work on this.

$$\displaystyle \sqrt{\frac{x-1}{x^5}} =\sqrt{\frac{1}{x^4}\frac{x-1}{x}}=\frac{1}{x^2}\sqrt{\frac{x-1}{x}}=\frac{1}{x^2}\sqrt{1-\frac{1}{x}}$$

Now let $$u=1-\frac{1}{x}$$, so $$du = \frac{1}{x^2}\,dx$$

$$\displaystyle \int \sqrt{\frac{x-1}{x^5}}\,dx=\int\frac{1}{x^2}\sqrt{1-\frac{1}{x}}\,dx=\int \sqrt{u}\,du=\frac{2}{3}u^\frac{3}{2}+C$$

Back substituting $$\displaystyle \int{\sqrt{\frac{x-1}{x^5}}}\,dx=\frac{2}{3}\left(1-\frac{1}{x}\right)^\frac{2}{3}+C$$

# Section 5.5 Substitution in Definite Integrals

Consider the chain rule:

$$! \frac{d}{dx}\bigg[F(g(x))\bigg]=F'(g(x))\cdot g'(x)$$

With respect to integration the chain rule looks like this:

$$! \int F'(g(x))\cdot g'(x)\, dx = F(g(x)) + C$$

Now consider a substitution, $$u=g(x)$$. Then $$\frac{du}{dx}=g'(x)$$ or in differential form $$du = g'(x)\,dx$$.

Applying this to the integral version of the chain rule, we get

$$! \int F'(u) \, du = F(u)+C$$

#### Example 1

$$! \frac{d}{dx}\bigg[\sin\left(x^2\right)\bigg]=\cos\left(x^2\right)\cdot 2x$$

This means that $$\sin\left(x^2\right)$$ is an antiderivative for $$\cos\left(x^2\right)\cdot 2x$$.

$$! \int \cos\left(x^2\right)\cdot 2x \, dx = \sin\left(x^2\right) + C$$

Applying the substitution $$u=x^2$$ and $$du=2x \, dx$$, we get

$$! \int \cos\left(u\right) \, du = \sin\left(u\right) + C$$

#### Other Examples

$$\displaystyle \int 3\cdot (3x+4)^7 \, dx$$

$$\displaystyle \int sqrt{5x-1} \, dx$$

$$\displaystyle \int \frac{4x^3}{x^4+1} \, dx$$

$$\displaystyle \int \sin^5\left( x/3\right) \cos\left( x/3\right) \, dx$$

$$\displaystyle \int_{0}^{\pi/2} \frac{2\sin(2\theta)}{4-\cos(2\theta)}\,d\theta$$

# Section 5.4

## The Fundamental Theorem of Calculus

#### Another Mean Value Theorem

$$\displaystyle f(c)=\frac{1}{b-a}\int_a^b f(x)dx$$ where $$f(c)$$ if the average value of $$f$$ on the interval $$[a,b]$$

#### The Fundamental Theorem of Calculus, part 1

If $$f$$ is continuous on $$[a,b]$$, then $$A(x)=\int_a^x f(t)dt$$ is continuous on $$[a,b]$$ and differentiable on $$(a,b)$$ and its derivative is f(x):

$$! A'(x)=\frac{d}{dx}\left[ \int_a^x f(t)dt \right]=f(x)$$

#### Example 1:

$$y=\int_1^x \frac{1}{t}dt$$ then $$\frac{dy}{dx} = \frac{1}{x}$$

The fundamental theorem is significant since it tells us that $$A(x)$$ is an antiderivative of $$f(x)$$

#### Example 2:

$$! A'(x)=\frac{d}{dx}\left[ \int_a^x f(t)dt \right]=f(x)$$

$$!\frac{d}{dx}\left[\sin(x)\right]=\cos(x)$$

$$!\frac{d}{dx}\left[\int_a^x \cos(t)dt\right] = \cos(x)$$

Since the right sides of each of these equations are the same, the left sides must be equal.

$$!\frac{d}{dx}\left[\sin(x)\right]=\frac{d}{dx}\left[\int_a^x \cos(t)dt\right]$$

We know from Calculus 1 that if two functions have the same derivative, then they must differ by no more than a constant.

$$\int_a^x \cos(t)dt – \sin(x) = C$$ or $$\int_a^x \cos(t)dt =\sin(x) – C$$

#### Proof of the fundamental theorem part 1:

Let $$A(x) = \int_a^x f(t)dt$$

Consider a small chunk of area $$\Delta A = A(x+h)-A(x)$$. We can also say that $$\Delta A \approx h\cdot f(x)$$

Now check out the rate of change of the area function $$A'(x)=\lim_{h\to 0}\frac{A(x+h)-A(x)}{h}$$

Notice that the numerator is also equal to $$h\cdot f(x)$$, fo in the limit $$A'(x)=f(x)$$. That is, not only does the boundary function drives the rate of change of the area, the boundary function is the rate of change of the area.