5.3 The Fundamental Theorem of Calculus

Using Riemann sums to calculate definite integrals can be tedious, as was seen in the previous section. In fact the technique shown in that section depended on the function being a low-degree polynomial, which obviously will not always be the case. Luckily there is a better way, involving antiderivatives, given by the following theorem:

Theorem 5.10

Fundamental Theorem of Calculus: Suppose that a function $f$ is differentiable on $\ival{a}{b}$. Then:

The function $A(x)$ defined on $\ival{a}{b}$ by

\[ A(x) ~=~ \int_a^x\,f(t)~\dt \]

is differentiable on $\ival{a}{b}$, and

\[ A'(x) ~=~ f(x) \]

for all $x$ in $\ival{a}{b}$.
If $F$ is an antiderivative of $f$ on $\ival{a}{b}$, i.e. $F'(x) = f(x)$ for all $x$ in $\ival{a}{b}$, then

\[ \int_a^b\,f(x)~\dx ~=~ F(b) ~-~ F(a) ~. \]

The function $A(x)$ in Part I of the theorem is sometimes called the area function because it represents the area under the curve $y=f(x)$ over the interval $\ival{a}{x}$, as shown in Figure 5.3.1 below.

tikzpicture[>=latex,every node/.style=font=] [fill=fillcolor] (3,1.5) to[out=180,in=0] (2,1) to[out=180,in=-45] (1,1.5) -- (1,0) -- (3,0) -- (3,1.5); [line width=1pt] (1,0) -- (1,1.5); [line width=1pt] (3,0) -- (3,1.5); [linecolor,line width=1.5pt] (4.2,1) to[out=140,in=-10] (3,1.5) to[out=180,in=0] (2,1) to[out=180,in=-45] (1,1.5); [above] at (2.5,1.5) y = f(x); [<->,black!60,line width=1pt,anchor=base] (0,2.3) node[above] y |- (5,0) node[right] x node[black,shift=(0,-0.4)] at (1,0) a node[black,shift=(0,-0.4)] at (3,0) x node[black,shift=(0,-0.4)] at (4.2,0) b; at (2,0.5) A(x); (1,1.5) circle (2.5pt); (4.2,1) circle (2.5pt); tikzpicture — Figure 5.3.1The area function $A(x) = \int_a^xf(t)\,\dt$

tikzpicture[>=latex,every node/.style=font=] [fill=fillcolor] (3,1.5) to[out=180,in=0] (2,1) -- (2,0) -- (3,0) -- (3,1.5); [line width=1pt] (2,0) -- (2,1); [line width=1pt] (3,0) -- (3,1.5); [linecolor,line width=1.5pt] (4.2,1) to[out=140,in=-10] (3,1.5) to[out=180,in=0] (2,1) to[out=180,in=-45] (1,1.5); [above] at (2.5,1.5) y = f(x); [<->,black!60,line width=1pt,anchor=base] (0,2.3) node[above] y |- (5,0) node[right] x node[black,shift=(0,-0.4)] at (1,0) a node[black,shift=(0,-0.4)] at (2,0) x node[black,shift=(0,-0.4)] at (3,0) x+ node[black,shift=(0,-0.4)] at (4.2,0) b; at (2.5,0.5) ; (1,1.5) circle (2.5pt); (4.2,1) circle (2.5pt); tikzpicture — Figure 5.3.2$\dA = A(x+\dx) - A(x)$

To prove Part I, assume that $f(x) \ge 0$ on $\ival{a}{b}$ as in Figure 5.3.1 (the proofs for $f(x)$ either negative or switching sign over $\ival{a}{b}$ are similar). The goal is to show that for any $x$ in $\ival{a}{b}$ the differential $\dA$ exists and equals $f(x)\,\dx$. First, $\dA = A(x+\dx) - A(x)$ is the area under the curve $y=f(x)$ over the interval $\ival{x}{x+\dx}$, as shown in Figure 5.3.2 above. By the Microstraightness Property the curve $y=f(x)$ is a straight line over the infinitesimal interval $\ival{x}{x+\dx}$, so $f$ must be either increasing, constant, or decreasing over that interval. The three possibilities are shown in Figure 5.3.3:

tikzpicture[>=latex,every node/.style=font=] [|<->|] (-0.4,0) -- (-0.4,4) node[left,midway] f(x); [|<->|] (1.9,0) -- (1.9,4) node[right,midway] f(x); [<->|] (1.9,4) -- (1.9,5.25) node[right,midway] ; [linecolor,fill=fillcolor] (0,0) -- (0,4) -- (1.5,5.25) -- (1.5,0) -- cycle; [linecolor,dashed] (0,4) -- (1.5,4); [linecolor] (1.3,4) -- (1.3,4.2) -- (1.5,4.2); [linecolor,line width=1.5pt] (0,4) -- (1.5,5.25) node[black,above left,midway] y=f(x); (0,4) circle (2.5pt); (1.5,5.25) circle (2.5pt); [below] at (0,0) x ; [below] at (1.5,0) x+ ; [|<->|] (0,-0.7) -- (1.5,-0.7) node[below,midway] ; [above] at (0,4) A; [above] at (1.5,5.25) B; [below left] at (1.5,4) C; tikzpicture — Figure 5.3.3The three possibilities for $\dA$

tikzpicture[>=latex,every node/.style=font=] [|<->|] (-0.4,0) -- (-0.4,4) node[left,midway] f(x); [linecolor,fill=fillcolor] (0,0) -- (0,4) -- (1.5,4) -- (1.5,0) -- cycle; [linecolor,line width=1.5pt] (0,4) -- (1.5,4) node[black,above,midway] y=f(x); (0,4) circle (2.5pt); (1.5,4) circle (2.5pt); [below] at (0,0) x ; [below] at (1.5,0) x+ ; [|<->|] (0,-0.7) -- (1.5,-0.7) node[below,midway] ; tikzpicture — Figure 5.3.3The three possibilities for $\dA$

In the case where $f$ is increasing over $\ival{x}{x+\dx}$, the infinitesimal area $\dA$ is the sum of the area of the rectangle of height $f(x)$ and width $\dx$ and the area of the right triangle $\triangle ABC$ shown in Figure 5.3.3(a). The area of $\triangle ABC$ is $\frac{1}{2}(\df)(\dx) = \frac{1}{2}f'(x)(\dx)^2 = 0$, so $\dA = f(x)\,\dx$.

In the case where $f$ is constant over $\ival{x}{x+\dx}$, the infinitesimal area $\dA$ is the area of the rectangle of height $f(x)$ and width $\dx$, as shown in Figure 5.3.3(b). So again, $\dA = f(x)\,\dx$.

In the case where $f$ is decreasing over $\ival{x}{x+\dx}$, the infinitesimal area $\dA$ is the sum of the area of the rectangle of height $f(x+\dx)$ and width $\dx$ and the area of the right triangle $\triangle ABC$ shown in Figure 5.3.3(c). Note that $\df < 0$ since $f$ is decreasing, and so the area of $\triangle ABC$ is $\frac{1}{2}(-\df)(\dx) = -\frac{1}{2}f'(x)(\dx)^2 = 0$. Thus,

\[ \dA ~=~ f(x+\dx)\,\dx ~=~ (f(x) + \df)\,\dx ~=~ f(x)\,\dx ~+~ f'(x)(\dx)^2 ~=~ f(x)\,\dx ~+~ 0 ~=~ f(x)\,\dx ~. \]

So in all three cases, $\dA = f(x)\,\dx$, and so $A'(x) = \frac{\dA}{\dx} = f(x)$, which shows that $A(x)$ is differentiable and has derivative $f(x)$. This proves Part I of the Fundamental Theorem of Calculus.$\quad\checkmark$

To prove Part II of the theorem, let $F(x)$ be an antiderivative of $f(x)$ over $\ival{a}{b}$. Since the are afunction $A(x) = \int_a^x f(x)\,\dx$ is also an antiderivative of $f(x)$ over $\ival{a}{b}$ by Part I of the theorem, then $A(x)$ and $F(x)$ differ by a constant $C$ over $\ival{a}{b}$. In other words:

\[ A(x) ~=~ F(x) ~+~ C \quad\text{for all $x$ in $\ival{a}{b}$} \]

By definition $A(a) = 0$, since it is the area under the curve over the interval $\ival{a}{a}$ of zero length. Thus,

\[ 0 ~=~ A(a) ~=~ F(a) ~+~ C \quad\Rightarrow\quad C ~=~ -F(a) \quad\Rightarrow\quad A(x) ~=~ F(x) ~-~ F(a) \quad\text{for all $x$ in $\ival{a}{b}$} \]

and so

\[ \int_a^b f(x)~\dx ~=~ A(b) ~=~ F(b) ~-~ F(a) \]

which proves Part II of the theorem.^[1] $\quad\checkmark$

Note: In some textbooks Part I is called the First Fundamental Theorem of Calculus and Part II is called the Second Fundamental Theorem of Calculus. The following notation provides a shorthand way of writing $F(b) - F(a)$:

Definition 5.7

\[ F(x)~\Biggr|_a^b ~=~ F(b) ~-~ F(a) \]

Example 5.12

Calculate $\displaystyle\int_1^2 x^2~\dx$.

Solution: Recall from Example 5.11 in the previous section that the integral equals $7/3$. In that example Riemann sums were used, but Part II of the Fundamental Theorem of Calculus makes the integral much easier to calculate. Since $F(x) = \frac{x^3}{3}$ is an antiderivative of $f(x) = x^2$, then

\[ \int_1^2 x^2~\dx ~=~ \frac{x^3}{3}~\Biggr|_1^{2} ~=~ \frac{2^3}{3} ~-~ \frac{1^3}{3} ~=~ \frac{7}{3} ~. \]

Note in the above example that any antiderivative of $f(x)=x^2$ could have been used, e.g. $F(x)= \frac{x^3}{3} + 5$. Notice that the constant 5 cancels out when evaluating $F(2)-F(1)$. So you do not need to add a generic constant $C$ to the antiderivative of $f(x)$ in a definite integral, as you would in an indefinite integral.

Example 5.13

Calculate $\displaystyle\int_0^{\pi} \sin\,x~\dx$.

Solution: Since $F(x) = -\cos\,x$ is an antiderivative of $f(x) = \sin\,x$, then

\[ \int_0^{\pi} \sin\,x~\dx ~=~ -\cos\,x~\Biggr|_0^{\pi} ~=~ -\cos\,\pi ~-~ (-\cos\,0) ~=~ -(-1) ~-~ (-1) ~=~ 2 ~. \]

Example 5.14

Calculate $\displaystyle\int_{-1}^{1} x^3~\dx$.

Solution: Since $F(x) = \frac{x^4}{4}$ is an antiderivative of $f(x) = x^3$, then

\[ \int_{-1}^{1} x^3~\dx ~=~ \frac{x^4}{4}~\Biggr|_{-1}^{1} ~=~ \frac{1^4}{4} ~-~ \frac{(-1)^4}{4} ~=~ \frac{1}{4} ~-~ \frac{1}{4} ~=~ 0~. \]

Example 5.14 is a special case of the following result for odd functions:

Theorem 5.11

If $f$ is an odd function, i.e. $f(-x) = -f(x)$ for all $x$, then

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

for all $a > 0$ such that $f$ is continuous on $\ival{-a}{a}$.

The idea is that since an odd function is symmetric around the origin, then the area between the curve and the $x$-axis over $\ival{0}{a}$ will cancel out the area between the curve and the $x$-axis over $\ival{-a}{0}$. Both areas are the same but one gets counted as positive and the other negative, as shown in Figure 5.3.4 below:

tikzpicture[>=latex,every node/.style=font=] [fill=fillcolor] (0,0) parabola (2,1.5) -- (2,0) -- cycle; [fill=fillcolor] (-2,-1.5) parabola[bend at end] (0,0) -- (-2,0) -- cycle; [linecolor] (2,0) -- (2,1.5); [linecolor] (-2,0) -- (-2,-1.5); [->,black!60,line width=1pt] (-2.5,0) -- (2.5,0) node[right] x; [->,black!60,line width=1pt] (0,-2) -- (0,2) node[above] y; [linecolor,line width=1.5pt] (-2,-1.5) parabola[bend at end] (0,0) parabola (2,1.5); [left] at (1.8,1.5) y=f(x); [below] at (2,0) a; [above] at (-2,0) -a; at (1.55,0.4) +; at (-1.55,-0.4) ; tikzpicture — Figure 5.3.4Odd function $f$ over $\ival{-a}{a}$

tikzpicture[>=latex,every node/.style=font=] [fill=fillcolor] (-2,3) parabola bend (0,0) (2,3) -- (2,0) -- (-2,0) -- cycle; [linecolor] (-2,0) -- (-2,3); [linecolor] (2,0) -- (2,3); [->,black!60,line width=1pt] (-2.5,0) -- (2.5,0) node[right] x; [->,black!60,line width=1pt] (0,-0.5) -- (0,3.5) node[above] y; [linecolor,line width=1.5pt] (-2,3) parabola bend (0,0) (2,3); [left] at (1.8,3) y=f(x); [below] at (2,0) a; [below] at (-2,0) -a; tikzpicture — Figure 5.3.5Even function $f$ over $\ival{-a}{a}$

By symmetry around the $y$-axis, a similar result holds for even functions (see Figure 5.3.5):

Theorem 5.12

If $f$ is an even function, i.e. $f(-x) = f(x)$ for all $x$, then

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

for all $a > 0$ such that $f$ is continuous on $\ival{-a}{a}$.

The following rules for definite integrals are a consequence of the corresponding rules for indefinite integrals:

Theorem 5.13

Let $f$ and $g$ be continuous functions on $\ival{a}{b}$ and let $k$ be a constant. Then:

$\displaystyle\int_a^b\,k\;f(x)~\dx ~=~ k\,\displaystyle\int_a^b\,f(x)~\dx$
$\displaystyle\int_a^b\,(f(x) + g(x))~\dx ~=~ \displaystyle\int_a^b\,f(x)~\dx ~+~ \displaystyle\int_a^b\,g(x)~\dx$
$\displaystyle\int_a^b\,(f(x) - g(x))~\dx ~=~ \displaystyle\int_a^b\,f(x)~\dx ~-~ \displaystyle\int_a^b\,g(x)~\dx$

The following results for definite integrals are a consequence of the Fundamental Theorem of Calculus:

Theorem 5.14

Let $f$ be a continuous function on $\ival{a}{b}$ and suppose that $a < c < b$. Then:

$\displaystyle\int_a^a\,f(x)~\dx ~=~ 0$
$\displaystyle\int_b^a\,f(x)~\dx ~=~ -\displaystyle\int_a^b\,f(x)~\dx$
$\displaystyle\int_a^b\,f(x)~\dx ~=~ \displaystyle\int_a^c\,f(x)~\dx ~+~ \displaystyle\int_c^b\,f(x)~\dx$

For example, if $F(x)$ is an antiderivative of $f(x)$ on $\ival{a}{b}$, then

\[ \int_a^c f(x)~\dx ~+~ \int_c^b f(x)~\dx ~=~ (F(c) - F(a)) ~+~ (F(b) - F(c)) ~=~ F(b) ~-~ F(a) ~=~ \int_a^b f(x)~\dx \]

which proves rule (3).

The following result is a consequence of Part I of the Fundamental Theorem of Calculus along with the Chain Rule:

Theorem 5.15

Chain Rule for integrals: Let $f$ be a continuous function on an interval $I$ containing $x=a$, and let $g(x)$ be a differentiable function on $I$. If

\[ F(x) ~=~ \int_a^{g(x)}\,f(t)~\dt \quad\text{for all $x$ in $I$} \]

then $F'(x) = f(g(x)) \cdot g'(x)$ for all $x$ in $I$.

Example 5.15

Let $F(x) = \displaystyle\int_0^{x^2} e^{-t^2}~\dt$ for all $x > 0$. Find $F'(x)$.

Solution: By the Chain Rule for integrals, with $f(t) = e^{t^2}$ and $g(x) = x^2$:

\[ F'(x) ~=~ f(g(x))\cdot g'(x) ~=~ e^{-(x^2)^2} \cdot (2x) ~=~ 2x\,e^{-x^4} \]

Exercises

For Exercises 1-12, evaluate the given definite integral.

$\displaystyle\int_{0}^{1}\, x^2 ~\dx$
$\displaystyle\int_{-1}^{1}\, x^2 ~\dx$
$\displaystyle\int_{0}^{1}\, x^3 ~\dx$
$\displaystyle\int_{-1}^{1}\, (x^2 + 3x - 4) ~\dx$

4
$\displaystyle\int_{1}^{2}\, \frac{1}{x^2} ~\dx\vphantom{\displaystyle\int_{0}^{\pi/2}}$
$\displaystyle\int_{2}^{3}\, \frac{1}{x^3} ~\dx\vphantom{\displaystyle\int_{0}^{\pi/2}}$
$\displaystyle\int_{0}^{\pi/2}\, \cos\,x ~\dx$
$\displaystyle\int_{0}^{1}\, e^x ~\dx\vphantom{\displaystyle\int_{0}^{\pi/2}}$

4
$\displaystyle\int_{-1}^{1}\, 2e^x ~\dx\vphantom{\dfrac{x^3\,e^{x^2}}{\cos\,2x}}$
$\displaystyle\int_{-\pi}^{\pi}\, \sin\,x ~\dx\vphantom{\dfrac{x^3\,e^{x^2}}{\cos\,2x}}$
$\displaystyle\int_{0}^{4}\, \sqrt{x} ~\dx\vphantom{\dfrac{x^3\,e^{x^2}}{\cos\,2x}}$
$\displaystyle\int_{-2}^{2}\, \dfrac{x^3\,e^{x^2}}{\cos\,2x} ~\dx$

2
Show that $\ln\,x ~=~ \displaystyle\int_{1}^{x}\, \frac{1}{t} ~\dt$ for all $x > 0$.
Show that $\displaystyle\int_b^a\,f(x)~\dx ~=~ -\displaystyle\int_a^b\,f(x)~\dx$.
Given $f(x) ~=~\displaystyle\int_{1}^{x}~\dfrac{t}{\sqrt{t^4 + 1}}~\dt~$, find $f'(3)$ and $f'(-2)$.

Prove the Chain Rule for integrals.
Explain why for any continuous function $f$ on $\ival{a}{b}$, $\ABS{\displaystyle\int_a^b\,f(x)~\dx} ~\le~ \displaystyle\int_a^b\,\abs{f(x)}~\dx ~.$
Explain why if $f(x) \le g(x)$ on $\ival{a}{b}$ then $\displaystyle\int_a^b\,f(x)~\dx ~\le~ \displaystyle\int_a^b\,g(x)~\dx ~.$

Show that if $f(x)$ is continuous on $\ival{a}{b}$ then there is a number $c$ in $(a,b)$ such that

\[ \int_a^b f(x)~\dx = f(c) \cdot (b-a) ~. \]
Let $f(t)$ be a continuous function for all $t \ge 0$, and for each $x \ge 0$ define a function $g(x)$ by

\[ g(x) ~=~ \displaystyle\int_0^x ~(x-t)\,f(t)~\dt ~. \]

Show that $g'(x) ~=~ \displaystyle\int_0^x ~f(t)~\dt ~$ for all $x \ge 0$.
Show that for all $x > 0$,

\[ \int_0^x \frac{\dt}{1+t^2} ~+~ \int_0^{1/x} \frac{\dt}{1+t^2} \]

is independent of $x$.

The theorem can be proved for the weaker condition that $f$ is merely continuous on $\ival{a}{b}$. See p.173-175 in Parzynski, W.R. and P.W. Zipse, Introduction to Mathematical Analysis, New York: McGraw-Hill, Inc., 1982. ↩