The Second Fundamental Theorem of Calculus

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The accumulation of a rate is given by the change in the amount.

There is a another common form of the Fundamental Theorem of Calculus. In this form, the Fundamental Theorem of Calculus will serve as our “shortcut formula” for calculating definite integrals.

Second Fundamental Theorem of Calculus Let $f$ be continuous on $[a,b]$. If $F$ is any antiderivative of $f$, then

\[ \int _a^b f(x)\d x = F(b)-F(a). \]

Call $G$ the accumulation function given by

\[G(x) = \int _a^x f(t) \d t.\]

By the First Fundamental Theorem of Calculus, $G$ is an antiderivative of $f$. We see that both:

\begin{align*} G(a) &= \int _a^a f(t) \d t = 0, \textrm { and } \\ G(b) &= \int _a^b f(x) \d x. \end{align*}

Fix $F$ as any antiderivative of $f$. Since $F$ and $G$ have the same derivative, they differ by a constant. That is, there is a constant $C$ such that for any $x$ in the interval $[a, b]$, $G(x) = F(x)+C$.

Then:

\begin{align*} \int _a^b f(x) \d x &= G(b) = G(b) - 0\\ &= G(b)-G(a)\\ &=( F(b)+C ) - ( F(a) + C )\\ &= F(b) - F(a). \end{align*}

From this you should see that the two versions of the Fundamental Theorem are very closely related. In reality, the two forms are equivalent, just differently stated. Hence people often simply call them both “The Fundamental Theorem of Calculus.” One way of thinking about the Second Fundamental Theorem of Calculus is:

This could be read as:

The accumulation of a rate is given by the change in the amount.

When we compute a definite integral, we first find an antiderivative and then evaluate at the limits of integration. It is convenient to first display the antiderivative and then evaluate. A special notation is often used in the process of evaluating definite integrals using the Fundamental Theorem of Calculus. Instead of explicitly writing $F(b)-F(a)$, we often write

\[ \eval {F(x)}_a^b \]

meaning that one should evaluate $F(x)$ at $b$ and then subtract $F(x)$ evaluated at $a$

\[ \eval {F(x)}_a^b = F(b)-F(a). \]

Let us examine the First and Second Fundamental Theorems together:

\begin{align*} \ddx \int _a^x f(t) \d t &= f(x), \textrm { (First Fundamental Theorem of Calculus)}\\ \int _a^b \ddx f(x) \d x &= \eval {f(x)}_a^b, \textrm { (Second Fundamental Theorem of Calculus)} \end{align*}

Together, they tell us how the definite integral and derivative interact with one-another.

Let’s see some examples of the fundamental theorem in action.

Compute:

\[ \int _{-2}^2 x^3\d x \]

We start by finding an antiderivative of $x^3$. One possible choice is $\frac {x^{4}}{4}$. (We can verify this by differentiating: $\ddx \frac {x^4}{4} = x^3$.) Notice that we do not need the “$+C$” term here. The Second Fundamental Theorem of Calculus says we can use any antiderivative of $x^3$. We will almost always use the one with $C=0$ when evaluating a definite integral.

\begin{align*} \int _{-2}^2 x^3 \d x &=\eval {\frac {x^4}{4}}_{-2}^2\\ &= \frac {2^4}{4} - \frac {(-2)^4}{4}\\ &= 0. \end{align*}

Compute:

\[ \int _0^\pi \sin (\theta ) \d \theta \]

We start by finding an antiderivative of $\sin (\theta )$. A correct choice is $-\cos (\theta )$, which one could verify this by taking the derivative. Then

\begin{align*} \int _0^\pi \sin (\theta ) \d \theta &= \eval {-\cos (\theta )}_0^\pi \\ &=\answer [given]{-\cos (\pi )} - (-\cos (0))\\ &=\answer [given]{2}. \end{align*}

This is interesting: It says that the area under one “hump” of a sine curve is $2$.

Compute:

\[ \int _0^5 e^t\d t \]

We start by finding an antiderivative of $e^t$. A correct choice is $\answer [given]{e^t}$, one could verify this by taking the derivative. Hence

\begin{align*} \int _0^5 e^t \d t &= \eval {e^t}_0^5 \\ &= e^5-\answer [given]{e^0}\\ &= e^5-\answer [given]{1}. \end{align*}

Compute:

\[ \int _1^2\left (x^9 + \frac {1}{x}\right ) \d x \]

We start by finding an antiderivative of $x^9 + \frac {1}{x}$. A correct choice is $\frac {x^{10}}{10} + \ln (x)$, one could verify this by taking the derivative. Hence

\begin{align*} \int _1^2\left (x^9 + \frac {1}{x}\right ) \d x &= \eval {\frac {x^{10}}{10} + \ln (x)}_1^2 \\ &= \frac {2^{10}}{10} + \ln (2) - \answer [given]{\frac {1}{10}}. \end{align*}

1 Understanding motion with the Fundamental Theorem of Calculus

We know that

The derivative of a position function is a velocity function.
The derivative of a velocity function is an acceleration function.

Now consider definite integrals of velocity and acceleration functions. Specifically, if $v(t)$ is a velocity function, what does $\displaystyle \int _a^b v(t)\d t$ mean?

The Second Fundamental Theorem of Calculus states that

\[ \int _a^b v(t)\d t = V(b) - V(a), \]

where $V(t)$ is any antiderivative of $v(t)$. Since $v(t)$ is a velocity function, we can choose $V(t)$ to be the position function. Then, $V(b) - V(a)$ measures a change in position, or displacement over the time interval $[a,b]$.

A ball is thrown straight up with velocity given by $v(t) = -32t+20 \unit {ft/s}$, where $t$ is measured in $\unit {seconds}$. Find, and interpret, $\displaystyle \int _0^1 v(t)\d t$.

Using the Second Fundamental Theorem of Calculus, we have

\begin{align*} \int _0^1 v(t)\d t &= \int _0^1 (-32t+20)\d t \\ &= \eval {\answer [given]{-16t^2 + 20t}}_0^1 \\ &= 4. \end{align*}

Thus if a ball is thrown straight up into the air with velocity

\[ v(t) = \answer [given]{-32t+20}, \]

the height of the ball, $1 \unit {second}$ later, will be $4 \unit {feet}$ above the initial height. Note that the ball has traveled much farther. It has gone up to its peak and is falling down, but the difference between its height at $t=0$ and $t=1$ is $4 \unit {ft}$.

Now we know that to solve certain kinds of problems, those that involve accumulation of some form, we “merely” find an antiderivative and substitute two values and subtract. Unfortunately, finding antiderivatives can be quite difficult. While there are a small number of rules that allow us to compute the derivative of any common function, there are no such rules for antiderivatives. There are some techniques that frequently prove useful, but we will never be able to reduce the problem to a completely mechanical process.