# Calculus: Integrals III

## The Fundamental Theorem of Calculus

How can one compute the exact area under a curve? Definite integrals are defined in terms of infinite sums, although they can be quite tedious to calculate. There must be another way! Well, there is. It's called the Fundamental Theorem of Calculus. It's fundamental because it connects the two main concepts in calculus, derivatives and integrals, tightly together.

**Fundamental Theorem of Calculus, Part 1:** Let $f(x)$ be a continuous function defined over some interval $[a,b]$. Define the function $F(x)$ as

$$F(x)=\displaystyle\int\limits_{a}^{x} f(t) \, dt $$

Then $F(x)$ is continuous on $[a,b]$, differentiable on $(a,b)$, and has the property that, for all $x \in [a,b]$,

$$F'(x) = f(x)$$

The function $F(x)$ is called the **antiderivative** of $f(x)$, or more informally "the integral" of $f(x)$. It is the function whose derivative is $f(x)$. It turns out that there isn't just one such function, but a whole family of functions. The derivative of a constant is zero, so the antiderivative of a function can have any constant term. For example, both $g(x)=x^2+3$ and $h(x)=x^2-9$ are antiderivatives of $f(x)=2x$. In fact, the constant term of an antiderivative is arbitrary. This arbitrary constant is usually denoted with the letter $C$, like so:

$$\displaystyle\int 2x \,dx = x^2 + C$$

The above notation, lacking the $a$ and $b$ at the tips of the integral sign, is for an **indefinite integral**, which provides the antiderivative of the integrand but does not use it to calculate an area. Just like you can take a derivative without necessarily using it to compute the instantaneous rate of change for some point on a function, so too can you take an integral without having to compute an area.

The plot thickens, though, as the theorem has a second part:

**Fundamental Theorem of Calculus, Part 2:** Letting $f(x)$ and $F(x)$ be as before, then

$$\int\limits_{a}^{b} f(x) \, dx = F(b) - F(a)$$

This is nifty. In order to evaluate a definite integral, just find the antiderivative, evaluate it at the two ends of the region of integration, and subtract one from the other. Of course, this entails taking the antiderivative. Which one? It turns out that it doesn't matter - the arbitrary constants cancel one another out in the subtraction. In fact, this arbitrary constant is by convention left out when taking definite integrals, since there's no point in including it.

A little intuition can go a long way. Integrating amounts to trying to "undo" a derivative. Sometimes this is easy, but many functions require more complicated solutions. Some can't even be integrated analytically at all. A little geometric intuition here goes a long way: Differentiating amounts to *dividing* a $y$ value by an $x$ value to get a rate. In contrast, integrating amounts to *multiplying* a $y$ value by an $x$ value to get an area.

If any of this is confusing, doing practice problems is the best way to see how things work.

## Problems

Differentiate the following functions:

- $f(x) = \displaystyle\int\limits_{3}^{x} 2t^2 - 1 \, dt$
- $g(x) = \displaystyle\int\limits_{9}^{x} t^4 - t^3 - t^2 + 10 \, dt$
- $h(x) = \displaystyle\int\limits_{-1}^{x} \cos(t) \, dt$
- $p(t) = \displaystyle\int\limits_{1}^{t} \dfrac{4}{3x} \, dx$
- $q(r) = \displaystyle\int\limits_{0}^{r} e^{14x} \, dx$

The Fundamental Theorem of Calculus makes taking these derivatives easy as pie:

- $f'(x) = 2x^2 - 1$
- $g'(x) = x^4 - x^3 - x^2 + 10$
- $h'(x) = \cos(x)$
- $p'(t) = \dfrac{4}{3t}$
- $q'(r) = e^{14r}$

Use the chain rule to evaluate the following derivative: $\dfrac{d}{dx}\displaystyle\int\limits_{1}^{\sin(x)} \dfrac{1}{t^2+1} \, dt$

Notice that the integral is not of the form specific in the Fundamental Theorem of Calculus. Let's substitute $u=\sin(x)$ into the equation. We need to remember to multiply by $\dfrac{du}{dx}=\cos(x)$ to satisfy the chain rule:

$ \dfrac{d}{dx}\displaystyle\int\limits_{1}^{\sin(x)} \dfrac{1}{t^2+1} \, dt = \dfrac{d}{du}\left(\displaystyle\int\limits_{1}^{u} \dfrac{1}{t^2+1} \, dt\right) \dfrac{du}{dx} \\ \dfrac{d}{dx}\displaystyle\int\limits_{1}^{\sin(x)} \dfrac{1}{t^2+1} \, dt = \dfrac{1}{u^2+1} \dfrac{du}{dx} \\ \dfrac{d}{dx}\displaystyle\int\limits_{1}^{\sin(x)} \dfrac{1}{t^2+1} \, dt = \dfrac{1}{\sin^2(x)+1} \cos(x) \\ $

Let $F(x) = \displaystyle\int\limits_{3}^{x} 2t^2 + 2t + 2 \, dt$. Find $F'(2) + F''(4)$.

**Step 1:**Find the derivatives:By the Fundamental Theorem of Calculus, $F'(x) = 2x^2 + 2x + 2$.

Differentiating $F'(x)$ again gives $F''(x) = 4x + 2$.

**Step 2:**Calculate the sum:$ F'(2) + F''(4) = \left(2(2)^2 + 2(2) + 2\right) + \left(4(4) + 2\right) \\ F'(2) + F''(4) = \left(8 + 4 + 2\right) + \left(16 + 2\right) \\ F'(2) + F''(4) = 32 \\ $

Evaluate: $\displaystyle\int\limits_{o}^{\pi} \cos(\theta) \, d\theta$

Remember that $\frac{d}{d\theta}\sin(\theta) = \cos(\theta)$, then apply the second part of the Fundamental Theorem of Calculus:

$ \displaystyle\int\limits_{o}^{\pi} \cos(\theta) \, d\theta = \sin(\pi) - \sin(0) \\ \displaystyle\int\limits_{o}^{\pi} \cos(\theta) \, d\theta = 0 - 0 \\ \displaystyle\int\limits_{o}^{\pi} \cos(\theta) \, d\theta = 0 \\ $

Let $G(x) = \displaystyle\int\limits_{x}^{4} \sin(t) \, dt$. Find $G'(x)$.

Note that the form of the integral is not exactly what the Fundamental Theorem of Calculus requires. Namely, the upper bound should be $x$, but here it is the lower bound.

Remember that $\displaystyle\int\limits_{a}^{b} f(t) \, dt = - \displaystyle\int\limits_{b}^{a} f(t) \, dt$. Use this fact to rearrange the integral to out liking:

$ G(x) = \displaystyle\int\limits_{x}^{4} \sin(t) \, dt \\ G(x) = -\displaystyle\int\limits_{4}^{x} \sin(t) \, dt \\ G'(x) = -\dfrac{d}{dx} \displaystyle\int\limits_{4}^{x} \sin(t) \, dt \\ G'(x) = -\sin(x) \\ $

Let $H(x) = \displaystyle\int\limits_{2x}^{5x} \dfrac{1}{t^2+1} \, dt$. Use the properties of definite integrals and the Fundamental Theorem of Calculus to find $H'(x)$.

Remember that $\displaystyle\int\limits_{a}^{b} f(x) \, dx = \displaystyle\int\limits_{a}^{c} f(x) \, dx + \displaystyle\int\limits_{c}^{b} f(x) \, dx$. Letting $c$ be any arbitrary constant, we can make the following manipulation:

$ H(x) = \displaystyle\int\limits_{2x}^{5x} \dfrac{1}{t^2+1} \, dt \\ H(x) = \displaystyle\int\limits_{2x}^{c} \dfrac{1}{t^2+1} \, dt + \displaystyle\int\limits_{c}^{5x} \dfrac{1}{t^2+1} \, dt \\ H(x) = - \displaystyle\int\limits_{c}^{2x} \dfrac{1}{t^2+1} \, dt + \displaystyle\int\limits_{c}^{5x} \dfrac{1}{t^2+1} \, dt \\ H(x) = \displaystyle\int\limits_{c}^{5x} \dfrac{1}{t^2+1} \, dt - \displaystyle\int\limits_{c}^{2x} \dfrac{1}{t^2+1} \, dt \\ $

Now we need to make a substitution. Let $p = 5x$ and let $q = 2x$

$ H(x) = \displaystyle\int\limits_{c}^{p} \dfrac{1}{t^2+1} \, dt - \displaystyle\int\limits_{c}^{q} \dfrac{1}{t^2+1} \, dt \\ $

Now use the chain rule and the Fundamental Theorem of Calculus to differentiate to find $H'(x)$:

$ H'(x) = \dfrac{d}{dp}\left(\displaystyle\int\limits_{c}^{p} \dfrac{1}{t^2+1} \, dt\right)\dfrac{dp}{dx} - \dfrac{d}{dq}\left(\displaystyle\int\limits_{c}^{q} \dfrac{1}{t^2+1} \, dt\right)\dfrac{dq}{dx} \\ H'(x) = \dfrac{1}{p^2+1} \dfrac{dp}{dx} - \dfrac{1}{q^2+1} \dfrac{dq}{dx} \\ H'(x) = \dfrac{1}{(5x)^2+1} (5) - \dfrac{1}{(2x)^2+1} (2) \\ H'(x) = \dfrac{5}{25x^2+1} - \dfrac{2}{4x^2+1} \\ $

Evaluate: $\displaystyle\int\limits_{0}^{4} 2^{x} \, dx$

$ \displaystyle\int\limits_{0}^{4} 2^{x} \, dx = \displaystyle\int\limits_{0}^{4} e^{\ln(2)x} \, dx $

Remember that $\frac{d}{dx}e^{ax}=a e^{ax}$. We can then reason that $\displaystyle\int e^{ax} \, dx = \dfrac{1}{a} e^{ax} + C$. Differetiating confirms that this is the case. Using this formula, we can now take the following steps:

$ \displaystyle\int\limits_{0}^{4} 2^{x} \, dx = \left. \dfrac{1}{\ln(2)}e^{\ln(2)x} \right|_{0}^{4} \\ \displaystyle\int\limits_{0}^{4} 2^{x} \, dx = \left. \dfrac{1}{\ln(2)}2^{x} \right|_{0}^{4} \\ \displaystyle\int\limits_{0}^{4} 2^{x} \, dx = \dfrac{1}{\ln(2)}2^{4} - \dfrac{1}{\ln(2)} 2^0 \\ \displaystyle\int\limits_{0}^{4} 2^{x} \, dx = \dfrac{1}{\ln(2)}16 - \dfrac{1}{\ln(2)} 1 \\ \displaystyle\int\limits_{0}^{4} 2^{x} \, dx = \dfrac{15}{\ln(2)} \\ $

Solve for $a$: $\displaystyle\int\limits_{a}^{10} 2x \, dx = 99$

Remember that the indefinite integral of $2x$ is $x^2+C$. Now apply the second part of the Fundamental Theorem of Calculus:

$\displaystyle\int\limits_{a}^{10} 2x \, dx = 99 \\ 10^2 - a^2 = 99 \\ 100 - a^2 = 99 \\ 1 - a^2 = 0 \\ a^2 = 1 \\ a = \pm 1 $

Let $b(9) = b(4) + 12$. Calculate $\displaystyle\int\limits_{4}^{9} b'(x) \, dx$

$ \displaystyle\int\limits_{4}^{9} b'(x) \, dx = b(9) - b(4) \\ \displaystyle\int\limits_{4}^{9} b'(x) \, dx = \left(b(4)+12\right) - b(4) \\ \displaystyle\int\limits_{4}^{9} b'(x) \, dx = 12 \\ $

When computing a definite integral using the theorem of calculus, the arbitrary constant of integration, $C$, is always left out. Show that this constant always cancels out, and that it is therefore okay to always omit it.

Let the antiderivative of $f(x)$ be of the form $F(x)=g(x)+C$, where $C$ is the arbitrary constant of integration and $g(x)$ consists of all other terms in the antiderivative. The second part of the Fundamental Theorem of Calculus says that

$\displaystyle\int\limits_{a}^{b} f(x) = F(b) - F(a)$

Substituting our special form of the antiderivatives gives us the following:

$ \displaystyle\int\limits_{a}^{b} f(x) = \left(g(b) + C\right) - (\left(g(a) + C\right) \\ \displaystyle\int\limits_{a}^{b} f(x) = g(b) - g(a) \\ $

We can see that the arbitrary constant of integration cancels itself out when the second term is subtracted from the first, and that the result of the definite integral does not in any way depend on it.