The First Fundamental Theorem of Calculus

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\d }[0]{\,d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[1]{{\overset {\rightharpoonup }{#1}}} \newcommand {\vec }[1]{{\overset {\boldsymbol {\rightharpoonup }}{\mathbf {#1}}}\hspace {0in}} \newcommand {\point }[1]{\left (#1\right )} \newcommand {\pt }[1]{\mathbf {#1}} \newcommand {\Lim }[2]{\lim _{\point {#1} \to \point {#2}}} \DeclareMathOperator {\proj }{\mathbf {proj}} \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\vec {\boldsymbol {\l }}} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

The rate that accumulated area under a curve grows is described identically by that curve.

Accumulation functions

While the definite integral computes a signed area, which is a fixed number, there is a way to turn it into a function.

Given a function $f$ that is continuous on $[a,b]$ , an accumulation function, for $a\le x\le b$ , is defined by $F(x) = \int _a^x f(t) \d t.$

One thing that you might notice is that an accumulation function seems to have two variables: $x$ and $t$ . Let’s see if we can explain this. Consider the following graph:

An accumulation function $F$ measures the signed area in the region $[a,x]$ between $f$ and the $t$ -axis. Hence $t$ is playing the role of a “place-holder” that allows us to evaluate $f$ . On the other hand, $x$ is the specific number that we are using to bound the region that will determine the area between $f$ and the $t$ -axis, and hence the value of $F$ .

Given $F(x) = \int _{-3}^x 4 \d t,$ what is $F(5)$ ? $F(5) = \answer {32}$

What is $F(-5)$ ? $F(-5) = \answer {-8}$

What is $F(-3)$ ? $F(-3) = \answer {0}$

Consider the following accumulation function for $f(x) = x^3$ . $F(x) = \int _{-1}^x t^3 \d t.$ Considering the interval $[-1,1]$ , where is $F$ increasing? Where is $F$ decreasing? When does $F$ have local extrema?

Below we can see a graph of $f$ along with the signed area measured by the accumulation function.

The accumulation function starts off at zero, and then as $x$ grows, $F$ is increasingdecreasing as the function accumulates negatively signed area.

However when $x>0$ , $F$ starts to accumulate positively signed area, and hence is . Thus $F$ is increasing on $(0,1)$ , decreasing on $(-1,0)$ and hence has a local minimum at $(0,0)$ .

Working with the accumulation function leads us to a question, what is $\int _a^x f(t) \d t$ when $x< a$ ? The general convention is that $\int _a^b f(t) \d t = -\int _b^a f(t) \d t.$ With this in mind, let’s consider one more example.

Consider the following accumulation function for $f(x) = x^3$ . $F(x) = \int _{-1}^x t^3 \d t.$ Where is $F$ increasing? Where is $F$ decreasing? When does $F$ have local extrema?

From our previous example, we know that $F$ is increasing on $(0,1)$ . Since $f$ continues to be positive at $t=1$ and beyond, $F$ is increasingdecreasing on $(0,\infty )$ . On the other hand, we know from our previous example that $F$ is increasingdecreasing on $(-1,0)$ .

For values to the left of $t=-1$ , $F$ is still decreasing, as less and less positively signed area is accumulated. Hence $F$ is increasing on $(0,\infty )$ , decreasing on $(-\infty ,0)$ and hence has an absolute minimum at $(0,0)$ .

The key point to take from these examples is that an accumulation function $F(x) = \int _a^x f(t) \d t$ is increasing precisely when $f$ is positive and is decreasing precisely when $f$ is negative. In short, it seems that $f$ is behaving in a similar fashion to $F'$ .

The First Fundamental Theorem of Calculus

Let $f$ be a continuous function on the real numbers and consider $F(x) = \int _a^x f(t)\d t.$ From our previous work we know that $F$ is increasing when $f$ is positive and $F$ is decreasing when $f$ is negative. Moreover, with careful observation, we can even see that $F$ is concave up when $f'$ is positive and that $F$ is concave down when $f'$ is negative. Thinking about what we have learned about the relationship of a function to its first and second derivatives, it is not too hard to guess that there must be a connection between $F'$ and the function $f$ . This is a good guess, check out our next theorem:

First Fundamental Theorem of Calculus Suppose that $f$ is continuous on the real numbers and let $F(x)=\int _a^x f(t)\d t.$ Then $F'(x)=f(x)$ .

The First Fundamental Theorem of Calculus says that an accumulation function of $f$ is an antiderivative of $f$ . Another way of saying this is:

This could be read as:

The rate that accumulated area under a curve grows is described identically by that curve.

Now that we are working with accumulation functions, let’s see what happens when we compose them with other functions.

Find the derivative $\ddx \int _2^{x^2} \ln t \d t.$

Consider $F(x) = \int _2^x \ln t \d t$ and set $h(x) = x^2$ . Now $\int _2^{x^2} \ln t \d t = F(h(x)).$ The First Fundamental Theorem of Calculus states that $F'(x) = \ln x$ . The chain rule gives us $\begin{align*} \ddx \int_2^{x^2} \ln t \d t&= \ddx F(h(x)) \\ &= F' (h(x))h'(x) \\ &= \ln (h(x)) h'(x) \\ &= \answer[given]{\ln (x^2) 2x}. \end{align*}$

Let’s practice this once more.

Find the derivative $\ddx \int _{\cos x}^5 t^3\d t.$

Consider $F(x) = \int _5^x t^3 \d t$ and set $h(x) = \answer [given]{\cos (x)}$ . Now $\int _{\cos x}^5 t^3\d t =- \int _{5}^{\cos x} t^3\d t = -F(h(x)).$ The First Fundamental Theorem of Calculus states that $F'(x) = x^3$ . The chain rule gives us $\begin{align*} \ddx\int_{\cos x}^5 t^3\d t &= \ddx (-F(h(x)))\\ &= -F'(h(x))h'(x)\\ &=-\left(h(x)\right)^3 h'(x)\\ &=\answer[given]{-\cos^3(x) (-\sin(x))}. \end{align*}$

Given the graph of a function $f$ on the interval $[-1,5]$ , sketch the graph of the accumulation function $F(x) = \int _{-1}^x f(t) \d t, \hspace {0.2in} -1\le x\le 5.$

First, we evaluate $F$ at some significant points. $F(-1) = \int _{-1}^{-1} f(t) \d t=\answer [given]{0}.$ $F(0) = \int _{-1}^{0} f(t) \d t=\answer [given]{-1.5}.$ $F(1) = \int _{-1}^{1} f(t) \d t=\answer [given]{-2}.$ $F(3) = \int _{-1}^{3} f(t) \d t=\answer [given]{0}.$ $F(4) = \int _{-1}^{4} f(t) \d t=\answer [given]{2}.$ $F(5) = \int _{-1}^{5} f(t) \d t=\answer [given]{4}.$

Since $F'(x)=f(x)$ , it follows that the function $F$ is increasing on the interval $\left (\answer [given]{1},\answer [given]{5}\right )$ and decreasing on the interval $\left (\answer [given]{-1},\answer [given]{1}\right ).$ Since the function $f$ , the derivative of $F$ , is increasing on $(-1,3)$ , $F$ is concave up on the interval $\left (\answer [given]{-1},\answer [given]{3}\right ).$ Since $f$ is constant on the interval $[3,5]$ , $F$ is linear on $[3,5]$ .

The function $F$ has a local and global minimum at $x=\answer [given]{1},$ and the global maximum at $x=\answer [given]{5}.$ Now we are ready to sketch the graph of $F$ , on the same set of axes as the graph of $f$ .

What if the function $f$ is the velocity function for an object moving along a straight line, i.e. $f(t)=v(t)$ , $a\le t\le b$ ?

What is the meaning of an accumulation function in that case?

We use different variables in that case, since we want $F$ to be a function of time, $t$ . With this adjustment, we define the accumulation function $F$ as follows. $F(t)=\int _a^t v(z)\d z.$ Since the integral $\int _a^t v(z)\d z$ gives the displacement of the object on the time interval $[a,t]$ , it follows that $s(t)-s(a)= \int _a^t v(z)\d z,$ where $s(t)$ gives the position of the object at the time $t$ . If we differentiate this equation with respect to $t$ , we get that $s'(t)=\dd t \int _a^t v(z)\d z.$ Since $s'(t)=v(t)$ , we have that $v(t)=\dd t \int _a^t v(z)\d z.$ This is the the First Fundamental Theorem of Calculus!