The Fundamental Theorem Of Calculus

The Fundamental Theorem of Calculus links the two branches of calculus: differentiating an integral with a variable upper limit gives back the integrand, and evaluating a definite integral reduces to finding an antiderivative and subtracting.

Formula	Name	Notes
$\frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)$	FTC, Part 1	$f$ continuous on $[a, b]$
$\int_{a}^{b} f (x) d x = F (b) - F (a)$	FTC, Part 2	$F$ is any antiderivative of $f$
$\int f (x) d x = \int_{a}^{x} f (t) d t + C$	Integral Form of Part 1	Explains one symbol for two ideas
${F (x) \|}_{a}^{b} = {[F (x)]}_{a}^{b} = F (b) - F (a)$	Evaluation Bar Notation	Brackets used when $F$ has several terms
$\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$	Breaking at a Jump	Works for jumps, fails for vertical asymptotes

So far, differential calculus and integral calculus have seemed to be two completely separate branches of mathematics. Differential calculus arose from constructing the tangent line to a curve and studying rates of change. Integral calculus arose from calculating the area under a curve. The Fundamental Theorem of Calculus connects these two branches. It also explains why we use $\int$ for both indefinite integrals and definite integrals.

The Area Function

When $f (t)$ is a known continuous function and a and b are two constants, the value of the integral $\int_{a}^{b} f (t) d t$ is a definite number. Hence, if we replace the upper limit b by a variable x, the definite integral becomes a function of x, say $A (x)$ . Graphically,

A (x) = \int_{a}^{x} f (t) d t = the net area between y = f (x) and the x -axis on [a, x] .

Note carefully the two different letters. Inside the integral, t is the dummy variable being swept from a up to the endpoint; x is the endpoint itself and is the variable of the function $A$ .

Shaded region under a curve from a to a variable point x defining the area function A of x — The area function $A (x) = \int_{a}^{x} f (t) d t$ .

Suppose x takes on an increment h. Then

A (x + h) = \int_{a}^{x + h} f (t) d t,

The same region extended from a to x plus h — Increasing the upper limit from $x$ to $x + h$ .

and the value of $A$ changes by

\begin{aligned} A(x+h)-A(x) &=\int_{a}^{x+h}f(t)\,dt-\int_{a}^{x}f(t)\,dt\\ &=\int_{a}^{x+h}f(t)\,dt+\int_{x}^{a}f(t)\,dt\\ &=\int_{x}^{x+h}f(t)\,dt &&{\small(\text{Property 4 of Section 7.2})} \end{aligned}

The thin sliver of area between x and x plus h, nearly a rectangle of height f of x and width h — For small $h$ , the sliver of new area is nearly a rectangle of height $f (x)$ and width $h$ .

For small h, that thin sliver of area is nearly a rectangle:

\int_{x}^{x + h} f (t) d t \approx f (x) h .

\frac{A (x + h) - A (x)}{h} \approx f (x) .

As $h \to 0$ , we can replace the approximation symbol by equality. That is,

\underbrace{\lim_{h\to0}\frac{A(x+h)-A(x)}{h}}_{=A'(x)}=f(x).

In other words, we have shown that

A'(x)=\frac{d}{dx}\int_{a}^{x}f(t)\,dt=f(x).

This is the Fundamental Theorem of Calculus.

The Fundamental Theorem of Calculus, Part 1. If $f$ is a continuous function on the interval $[a, b]$ , then the function $A (x)$ defined by

A (x) = \int_{a}^{x} f (t) d t (a \leq x \leq b)

is continuous on $[a, b]$ and differentiable on $(a, b)$ , and A'(x)=f(x).

The Fundamental Theorem of Calculus is often written as

\frac{d}{d x} \int_{a}^{x} f (t) d t = f (x) .

In words: accumulating a function and then differentiating the accumulation returns the original function. Accumulation and rate of change are inverse operations.

Since one antiderivative of $f (x)$ is given by

F (x) = \int_{a}^{x} f (t) d t,

and the general antiderivative of $f (x)$ , denoted by $\int f (x) d x$ , is simply $F (x) + C$ ,

\int f (x) d x = F (x) + C,

we can write

\int f (x) d x = \int_{a}^{x} f (t) d t + C .

This formula can be considered another representation of the first part of the Fundamental Theorem of Calculus.

The Second Part: How to Evaluate an Integral

The Fundamental Theorem of Calculus provides a simple method for evaluating definite integrals. The following theorem is sometimes called the second part of the Fundamental Theorem of Calculus.

Let

A (x) = \int_{a}^{x} f (t) d t .

We know from the first part that A'(x)=f(x); that is, $A (x)$ is an antiderivative of $f (x)$ . If $F (x)$ is any antiderivative of $f (x)$ on $[a, b]$ , then $F$ and $A$ differ by a constant, because two functions with the same derivative on an interval differ by a constant. So there is a constant $C$ such that

F (x) = A (x) + C .

Therefore

\begin{aligned} F(b)-F(a) &=\left(A(b)+C\right)-\left(A(a)+C\right)\\ &=A(b)-A(a)\\ &=\int_{a}^{b}f(t)\,dt-\underbrace{\int_{a}^{a}f(t)\,dt}_{=0}\\ &=\int_{a}^{b}f(t)\,dt. \end{aligned}

Notice that the unknown constant $C$ cancels, which is why any antiderivative will do.

The Fundamental Theorem of Calculus, Part 2. Suppose $f$ is a continuous function on the interval $[a, b]$ and $F$ is any antiderivative of $f$ ; that is, F'(x)=f(x). Then

\int_{a}^{b} f (x) d x = F (b) - F (a) .

A Useful Notation

The difference $F (b) - F (a)$ is often denoted by

{F (x) |}_{a}^{b} or {F (x)]}_{a}^{b} .

If $F$ has more than one term and there might be confusion about which terms are involved, we use

{[F (x)]}_{a}^{b} .

To emphasize that a and b are values for the variable x, we may write

{[F (x)]}_{x = a}^{x = b} .

The second part of the Fundamental Theorem of Calculus tells us how to compute the definite integral of a function:

First: Find the indefinite integral of the given function $f$ .

Second: Substitute into the indefinite integral first the upper limit and then the lower limit for the variable of integration, and subtract the second result from the first.

So the second part of the Fundamental Theorem of Calculus can be written as

\int_{a}^{b} f (x) d x = {[\int f (x) d x]}_{a}^{b} .

Now it makes sense why we use the same symbol $\int$ for both the indefinite integral and the definite integral.

In the indefinite integral, $\int$ has no upper and lower limits. The indefinite integral is a function (or a family of functions).

In the definite integral, $\int$ has upper and lower limits. The definite integral is a number.

Worked Examples

Find $\int_{1}^{3} x^{2} d x$ .

Solution

Because the indefinite integral of $x^{2}$ is

\int x^{2} d x = \frac{1}{3} x^{3} + C,

it follows from the second part of the Fundamental Theorem of Calculus that

\int_{1}^{3} x^{2} d x = {[\frac{1}{3} x^{3}]}_{1}^{3} = \frac{1}{3} (3^{3} - 1^{3}) = \frac{26}{3} .

This is exactly the same result we obtained with a Riemann sum in Section 7.1, in a fraction of the work.

Notice that $x^{3} / 3$ is one particular antiderivative of $x^{2}$ . Instead of $x^{3} / 3$ we could use $x^{3} / 3 + 2$ , or $x^{3} / 3 - \sqrt{2}$ , and so on; the constant does not affect the result. Let us verify this with $x^{3} / 3 - \sqrt{2}$ :

\begin{aligned} \left[\frac{1}{3}x^{3}-\sqrt{2}\right]_{1}^{3} &=\left(\frac{1}{3}\times3^{3}-\sqrt{2}\right)-\left(\frac{1}{3}\times1^{3}-\sqrt{2}\right)\\ &=\frac{1}{3}\times3^{3}-\frac{1}{3}\times1=\frac{26}{3}. \end{aligned}

To apply the second part of the Fundamental Theorem of Calculus, we just need any antiderivative (also called a particular integral function). Using the most general antiderivative is not necessary.

Find $\int_{0}^{π} \sin x d x$ .

Solution

\begin{aligned} \int_{0}^{\pi}\sin x\,dx &=-\cos x\big|_{0}^{\pi}\\ &=-\cos\pi-(-\cos0)\\ &=-(-1)-(-1)\\ &=2. \end{aligned}

Find $\int_{1}^{\sqrt{3}} \frac{1}{1 + x^{2}} d x$ .

Solution

Because

\int \frac{1}{1 + x^{2}} d x = \arctan x + C,

we have

\begin{aligned} \int_{1}^{\sqrt{3}}\frac{1}{1+x^{2}}\,dx &=\left.\arctan x\right|_{1}^{\sqrt{3}}\\ &=\arctan\sqrt{3}-\arctan1\\ &=\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}. \end{aligned}

[Some books use $\tan^{- 1}$ instead of $\arctan$ for the inverse of the tangent function.]

Find $\int_{e}^{8} \frac{1}{x} d x$ .

Solution

Because

\int \frac{1}{x} d x = \ln | x | + C,

we have

\begin{aligned} \int_{e}^{8}\frac{1}{x}\,dx &=\ln|x|\Big|_{e}^{8}\\ &=\ln8-\ln e\\ &=\ln2^{3}-\underbrace{\ln e}_{1}\\ &=3\ln2-1 &&{\small(\ln a^{b}=b\ln a)} \end{aligned}

Find $\int_{0}^{1} (x^{3} - 2 \sqrt{x} + \frac{4}{\sqrt{1 - x^{2}}} - 3 e^{2 x}) d x$ .

Solution

First, we need the indefinite integral of the integrand:

\begin{aligned} \int&\left(x^{3}-2\sqrt{x}+\frac{4}{\sqrt{1-x^{2}}}-3e^{2x}\right)dx\\ &=\int x^{3}\,dx-2\int x^{1/2}\,dx+4\int\frac{1}{\sqrt{1-x^{2}}}\,dx-3\int e^{2x}\,dx\\ &=\frac{1}{4}x^{4}-2\times\frac{2}{3}x^{3/2}+4\arcsin x-\frac{3}{2}e^{2x}. \end{aligned}

Therefore

\begin{aligned} \int_{0}^{1}&\left(x^{3}-2\sqrt{x}+\frac{4}{\sqrt{1-x^{2}}}-3e^{2x}\right)dx\\ &=\left[\frac{1}{4}x^{4}-\frac{4}{3}x^{3/2}+4\arcsin x-\frac{3}{2}e^{2x}\right]_{0}^{1}\\ &=\left(\frac{1}{4}-\frac{4}{3}+4\underbrace{\arcsin1}_{\pi/2}-\frac{3}{2}e^{2}\right)-\left(0-0+4\underbrace{\arcsin0}_{0}-\frac{3}{2}\underbrace{e^{0}}_{1}\right)\\ &=\frac{1}{4}-\frac{4}{3}+2\pi-\frac{3}{2}e^{2}+\frac{3}{2}\\ &=\frac{5}{12}+2\pi-\frac{3}{2}e^{2}. \end{aligned}

The arithmetic in the last step: $\frac{1}{4} - \frac{4}{3} + \frac{3}{2} = \frac{3 - 16 + 18}{12} = \frac{5}{12}$ .

Continuity Is Required

To use the second part of the Fundamental Theorem of Calculus, the integrand $f$ must be continuous between a and b: no holes, no jumps, and no vertical asymptotes are allowed.

Therefore, because $f (x) = 1 / x^{2}$ has a vertical asymptote at $x = 0$ , we cannot write

\xcancel{\int_{-2}^{1}\frac{1}{x^{2}}\,dx=\left[\frac{x^{-1}}{-1}\right]_{-2}^{1}=-\frac{1}{1}-\left(-\frac{1}{-2}\right)=-\frac{3}{2}.}

The result is not merely unjustified, it is absurd: the integrand is positive everywhere it is defined, so no correct value could be negative.

If there is a jump in the graph of $f$ at $x = c$ (with $a < c < b$ ), we can simply write

\int_{a}^{b} f (x) d x = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x,

and apply the second part of the Fundamental Theorem of Calculus to each of the integrals on the right-hand side.

A function with a jump discontinuity at x equals c, with the region split into two parts — If there is a jump, we can break up the integral into two parts and evaluate each separately: $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ .

This technique does not work if $f$ has a vertical asymptote at $x = c$ .

A function with a vertical asymptote between a and b where the Fundamental Theorem cannot be applied — If the graph of $f$ has a vertical asymptote between $x = a$ and $x = b$ , we cannot apply the Fundamental Theorem of Calculus to find the integral of $f$ from $a$ to $b$ .

Given

f(x)=\begin{cases} 2x-1 & -1\leq x\leq1\\ -x+1 & 1

find $\int_{- 1}^{2} f (x) d x$ .

Solution

We split at the jump and use a different formula on each piece:

\begin{aligned} \int_{-1}^{2}f(x)\,dx &=\int_{-1}^{1}f(x)\,dx+\int_{1}^{2}f(x)\,dx\\ &=\int_{-1}^{1}(2x-1)\,dx+\int_{1}^{2}(-x+1)\,dx\\ &=\left[x^{2}-x\right]_{-1}^{1}+\left[-\frac{1}{2}x^{2}+x\right]_{1}^{2}\\ &=(1-1)-\left(1-(-1)\right)+(-2+2)-\left(-\tfrac{1}{2}+1\right)\\ &=0-2+0-\frac{1}{2}\\ &=-\frac{5}{2}. \end{aligned}

Also, reading the areas directly off the graph gives the same result:

\begin{aligned} \int_{-1}^{2}f(x)\,dx &=-A_{1}+A_{2}-A_{3}\\ &=-\frac{1.5\times3}{2}+\frac{0.5\times1}{2}-\frac{1\times1}{2}\\ &=-\frac{5}{2}. \end{aligned}

Graph of the piecewise linear function f with the three triangular regions labelled — The graph of $f$ and the integral of $f$ from $- 1$ to $2$ .

Exercises

Evaluate:

(a) $\int_{0}^{2} (3 x^{2} - 4 x + 1) d x$

(b) $\int_{1}^{4} (\sqrt{x} + \frac{1}{\sqrt{x}}) d x$

(c) $\int_{0}^{π / 4} \sec^{2} x d x$

Answer

(a) $2$ (b) $\frac{20}{3}$ (c) $1$

Solution

(a) ${[x^{3} - 2 x^{2} + x]}_{0}^{2} = (8 - 8 + 2) - 0 = 2$ .

(b) ${[\frac{2}{3} x^{3 / 2} + 2 x^{1 / 2}]}_{1}^{4} = (\frac{16}{3} + 4) - (\frac{2}{3} + 2) = \frac{14}{3} + 2 = \frac{20}{3}$ .

(c) ${[\tan x]}_{0}^{π / 4} = 1 - 0 = 1$ .

Evaluate:

(a) $\int_{0}^{1} e^{3 x} d x$

(b) $\int_{1}^{e^{2}} \frac{d x}{x}$

(c) $\int_{0}^{1 / 2} \frac{d x}{\sqrt{1 - x^{2}}}$

Answer

(a) $\frac{e^{3} - 1}{3}$ (b) $2$ (c) $\frac{π}{6}$

Solution

(a) ${[\frac{1}{3} e^{3 x}]}_{0}^{1} = \frac{e^{3}}{3} - \frac{1}{3}$ .

(b) ${[\ln | x |]}_{1}^{e^{2}} = \ln e^{2} - \ln 1 = 2$ .

(c) ${[\arcsin x]}_{0}^{1 / 2} = \frac{π}{6} - 0$ .

Find $\int_{- 2}^{2} | x | d x$ by splitting the interval, and check your answer geometrically.

Answer

$4$

Solution

The formula for $| x |$ changes at $x = 0$ , so split there:

\begin{aligned} \int_{-2}^{2}|x|\,dx &= \int_{-2}^{0}(-x)\,dx+\int_{0}^{2}x\,dx\\ &= \left[-\frac{x^{2}}{2}\right]_{-2}^{0}+\left[\frac{x^{2}}{2}\right]_{0}^{2}\\ &= (0+2)+(2-0)=4. \end{aligned}

Geometrically the region is two triangles, each with base 2 and height 2, so the area is $2 \times \frac{2 \times 2}{2} = 4$ .

Note that $| x |$ is continuous, so no jump is involved; we split only because the formula, not the function, changes at the origin.

Let

f(x)=\begin{cases} x^{2} & 0\leq x\leq1\\ 2-x & 1

Evaluate $\int_{0}^{3} f (x) d x$ .

Answer

$\frac{1}{3}$

Solution

\begin{aligned} \int_{0}^{3}f(x)\,dx &= \int_{0}^{1}x^{2}\,dx+\int_{1}^{3}(2-x)\,dx\\ &= \left[\frac{x^{3}}{3}\right]_{0}^{1}+\left[2x-\frac{x^{2}}{2}\right]_{1}^{3}\\ &= \frac{1}{3}+\left[\left(6-\frac{9}{2}\right)-\left(2-\frac{1}{2}\right)\right]\\ &= \frac{1}{3}+\left(\frac{3}{2}-\frac{3}{2}\right)=\frac{1}{3}. \end{aligned}

The second integral vanishes because the line $y = 2 - x$ lies above the axis on $[1, 2]$ and below it on $[2, 3]$ , contributing equal and opposite areas $\frac{1}{2}$ and $- \frac{1}{2}$ .

Explain why the computation

\int_{- 1}^{1} \frac{d x}{x} = {[\ln | x |]}_{- 1}^{1} = \ln 1 - \ln 1 = 0

is invalid.

Answer

The integrand has a vertical asymptote at $x = 0$ , which lies inside the interval of integration, so Part 2 does not apply.

Solution

Part 2 of the Fundamental Theorem requires the integrand to be continuous on the whole closed interval $[- 1, 1]$ . Here $1 / x$ is undefined at $x = 0$ and has a vertical asymptote there, so the hypothesis fails.

The failure is not a technicality. Splitting at the asymptote gives $\int_{- 1}^{0} \frac{d x}{x}$ and $\int_{0}^{1} \frac{d x}{x}$ , and neither piece is a finite number; the "answer" 0 comes from cancelling two infinities. Contrast this with a jump discontinuity, where splitting the interval is legitimate because each piece has a finite integral.

Find the value of b $(b > 0)$ for which $\int_{0}^{b} (x^{2} - 2 x) d x = 0$ .

Answer

$b = 3$

Solution

\int_{0}^{b} (x^{2} - 2 x) d x = {[\frac{x^{3}}{3} - x^{2}]}_{0}^{b} = \frac{b^{3}}{3} - b^{2} .

Setting this equal to zero gives $b^{2} (\frac{b}{3} - 1) = 0$ , so $b = 0$ or $b = 3$ . Since $b > 0$ , we take $b = 3$ .

The interpretation: on $[0, 2]$ the parabola lies below the axis and on $[2, 3]$ above it, and $b = 3$ is exactly where the area above has grown to match the area below.

The velocity of a particle moving along a line is $v (t) = t^{2} - 4$ metres per second, for $0 \leq t \leq 3$ . Find

(a) the net displacement,

(b) the total distance travelled.

Answer

(a) $- 3$ m (b) $\frac{23}{3}$ m $\approx 7.67$ m

Solution

(a) The net displacement is the integral of the velocity:

\int_{0}^{3} (t^{2} - 4) d t = {[\frac{t^{3}}{3} - 4 t]}_{0}^{3} = (9 - 12) - 0 = - 3.

The particle ends 3 metres to the left of where it started.

(b) The total distance is the integral of $| v |$ . The velocity changes sign at $t = 2$ , so

\begin{aligned} \int_{0}^{3}|t^{2}-4|\,dt &= -\int_{0}^{2}(t^{2}-4)\,dt+\int_{2}^{3}(t^{2}-4)\,dt\\ &= -\left(\frac{8}{3}-8\right)+\left[(9-12)-\left(\frac{8}{3}-8\right)\right]\\ &= \frac{16}{3}+\left(-3+\frac{16}{3}\right)=\frac{16}{3}+\frac{7}{3}=\frac{23}{3}. \end{aligned}

This is the standard illustration of the difference between net and total: the integral of $v$ gives displacement, the integral of $| v |$ gives distance.

Show that for any function $f$ continuous on an interval containing a and b,

\int_{a}^{b} f (x) d x + \int_{b}^{a} f (x) d x = 0,

using Part 2 of the Fundamental Theorem rather than the definition.

Answer

Both integrals equal $F (b) - F (a)$ and $F (a) - F (b)$ , which are negatives of each other.

Solution

Let $F$ be any antiderivative of $f$ . By Part 2,

\int_{a}^{b} f (x) d x = F (b) - F (a), \int_{b}^{a} f (x) d x = F (a) - F (b) .

Adding the two gives $[F (b) - F (a)] + [F (a) - F (b)] = 0$ .

This confirms from the evaluation side what we adopted as a definition in Section 7.1, namely $\int_{b}^{a} f = - \int_{a}^{b} f$ : the two viewpoints agree, as they must.

Water flows into a tank at the rate $r (t) = 6 t - t^{2}$ litres per minute for $0 \leq t \leq 6$ . How much water enters the tank during the first 6 minutes, and at what time is the flow rate greatest?

Answer

$36$ litres; the rate is greatest at $t = 3$ minutes.

Solution

Since $r (t)$ is the rate of change of the volume $V (t)$ , Part 2 gives the net change directly:

V (6) - V (0) = \int_{0}^{6} (6 t - t^{2}) d t = {[3 t^{2} - \frac{t^{3}}{3}]}_{0}^{6} = (108 - 72) - 0 = 36 litres.

The flow rate is greatest where r'(t)=6-2t=0, that is at $t = 3$ , where $r (3) = 9$ litres per minute. Since r''=-2<0, this is a maximum.

This is the "net change" reading of the theorem: integrate a rate, get a total.

Suppose $F$ is an antiderivative of $f$ on $[a, b]$ and $G (x) = F (x) + 7$ . Show that $F$ and $G$ produce the same value for $\int_{a}^{b} f (x) d x$ , and explain in one sentence why this had to happen.

Answer

Both give $F (b) - F (a)$ ; the added constant cancels in the subtraction.

Solution

Using $G$ ,

G (b) - G (a) = [F (b) + 7] - [F (a) + 7] = F (b) - F (a),

which is exactly what $F$ gives.

The reason it had to happen: Part 2 subtracts two values of the same antiderivative, and any two antiderivatives of $f$ on an interval differ by a constant, so that constant is added once and subtracted once. This is why the phrase "any antiderivative" in the statement of the theorem is safe, and why we never bother writing $+ C$ when evaluating a definite integral.

The Area Function

The Second Part: How to Evaluate an Integral

A Useful Notation

Worked Examples

Continuity Is Required

Exercises

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