6 The Fundamental Theorem of Calculus

A clear progression through the Fundamental Theorem of Calculus, definite integrals, accumulation, net change, average value, and introductory applications of integration.

Antiderivatives and signed accumulation

Integration and differentiation undo one another in two complementary ways. Differentiation gives an instantaneous rate, while integration combines infinitely many small contributions into a total.

An is a function $F$ satisfying

F'(x)=f(x).

For instance,

\frac{d}{dx}\left(\frac{x^3}{3}\right)=x^2,

so $\frac{x^3}{3}$ is an of $x^2$ . The indefinite integral records the entire family:

\int x^2\,dx=\frac{x^3}{3}+C.

By contrast, a has limits:

\int_a^b f(x)\,dx.

It gives a numerical amount of signed accumulation. Contributions above the $x$ -axis are positive, while contributions below it are negative.

Takeaway: An indefinite integral describes a family of antiderivatives; a describes signed accumulation over a specified interval.

Differentiating accumulation functions

The connects accumulation to differentiation. Define

G(x)=\int_a^x f(t)\,dt.

If $f$ is continuous, then

G'(x)=f(x).

The variable $t$ is a dummy variable: it is used inside the integral so that $x$ can represent the variable upper limit.

For example, if

G(x)=\int_1^x\sqrt{1+t^4}\,dt,

then

G'(x)=\sqrt{1+x^4}.

No elementary of the integrand is needed.

When both limits depend on the variable, apply the chain rule to each limit. For

H(x)=\int_{x^2}^{\sin x}\cos(t^3)\,dt,

we obtain

H'(x)=\cos\bigl((\sin x)^3\bigr)\cos x-\cos\bigl((x^2)^3\bigr)(2x).

The upper-limit contribution is positive, and the lower-limit contribution is subtracted.

Takeaway: Differentiating an accumulation function returns its integrand, with chain-rule factors when the limits vary.

Evaluating definite integrals

The evaluates a . If $F$ is any of a continuous function $f$ on $[a,b]$ , then

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

Use this procedure:

Find an $F(x)$ .
Evaluate it at the upper limit.
Evaluate it at the lower limit.
Subtract the lower endpoint value from the upper endpoint value.

For example,

\int_1^3(2x^3-4x+1)\,dx

has

F(x)=\frac{x^4}{2}-2x^2+x.

Therefore,

\begin{aligned} \int_1^3(2x^3-4x+1)\,dx &=F(3)-F(1)\\ &=\left(\frac{81}{2}-18+3\right)-\left(\frac{1}{2}-2+1\right)\\ &=24. \end{aligned}

Do not add an arbitrary constant when evaluating a . If it were included, it would cancel:

(F(b)+C)-(F(a)+C)=F(b)-F(a).

Takeaway: Definite-integral evaluation is an endpoint subtraction, not an indefinite-integral answer with $C$ .

Rates, net change, and motion

The turns a known rate into the resulting change. If $Q'(t)$ is the rate of change of $Q(t)$ , then

Q(b)-Q(a)=\int_a^b Q'(t)\,dt.

Equivalently,

Q(b)=Q(a)+\int_a^b Q'(t)\,dt.

For motion, velocity is the rate of change of position. Suppose

v(t)=3t^2-4t

meters per second and $s(1)=5$ . The displacement from $t=1$ to $t=3$ is

\begin{aligned} \int_1^3v(t)\,dt &=\int_1^3(3t^2-4t)\,dt\\ &=\left[t^3-2t^2\right]_1^3\\ &=(27-18)-(1-2)\\ &=10\text{ m}. \end{aligned}

Thus,

s(3)=s(1)+10=15\text{ m}.

This integral gives displacement, which is signed. instead accumulates the magnitude of velocity:

\text{total distance}=\int_a^b|v(t)|\,dt.

Takeaway: To recover a final quantity, add the accumulated rate to the initial quantity; distinguish signed change from total magnitude.

Net area and geometric area

A gives net area, so cancellation can occur when a function crosses the horizontal axis. For

f(x)=x-1

on $[0,2]$ ,

\int_0^2(x-1)\,dx=\left[\frac{x^2}{2}-x\right]_0^2=0.

The negative triangular contribution on $[0,1]$ cancels the positive triangular contribution on $[1,2]$ . The total geometric area counts both regions positively:

\int_0^1(1-x)\,dx+\int_1^2(x-1)\,dx=\frac{1}{2}+\frac{1}{2}=1.

More generally, total area can be written as

\int_a^b|f(x)|\,dx,

or evaluated by splitting the interval wherever the function changes sign.

For two curves, identify the upper and lower functions. If $f(x)\geq g(x)$ on $[a,b]$ , then the is

A=\int_a^b\bigl(f(x)-g(x)\bigr)\,dx.

If the curves cross, split at their intersection points and reassess which function is on top.

Takeaway: Signed area allows cancellation; total area and require every geometric contribution to be counted with the correct positive height.

of a function

The of an integrable function on $[a,b]$ is

f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx.

It represents the constant height of a rectangle whose signed area equals the integral of $f$ . If $f$ is continuous, at least one point $c\in[a,b]$ satisfies

f(c)=f_{\text{avg}}.

For $f(x)=x^2$ on $[0,3]$ ,

\begin{aligned} f_{\text{avg}} &=\frac{1}{3-0}\int_0^3x^2\,dx\\ &=\frac{1}{3}\left[\frac{x^3}{3}\right]_0^3\\ &=3. \end{aligned}

The function reaches this when

x^2=3,

so the point in the interval is $x=\sqrt{3}$ .

Takeaway: Divide total signed accumulation by interval length to obtain the .

A strategy for integration applications

Integration applies whenever a total is built from a rate, density, or cross-sectional quantity. A reliable setup follows this sequence:

Identify the quantity being accumulated, such as displacement, volume, mass, or area.
Identify the rate, density, or cross-sectional function.
Choose limits that match the physical or geometric interval.
Write the and track units.
Find an when an explicit evaluation is possible.
Apply endpoint subtraction.
Check the sign and units of the result.
Split the interval when sign changes affect or total area.

Common applications include

Position from velocity:
$s(b)=s(a)+\int_a^b v(t)\,dt.$
Velocity from acceleration:
$v(b)=v(a)+\int_a^b a(t)\,dt.$
Accumulation from a rate $r(t)$ :
$\text{amount accumulated}=\int_a^b r(t)\,dt.$
Volume from a cross-sectional area $A(x)$ :
$V=\int_a^b A(x)\,dx.$

A negative result is not automatically an error: it may describe a decrease, negative displacement, or signed area. The interpretation depends on the quantity being modeled.

Final takeaway: First determine what is accumulating and why; then write the integral, evaluate it using the Fundamental Theorem of Calculus, and interpret the result with its sign and units.