09 Definite Integrals and the Fundamental Theorem

A progressive guide to definite integrals, signed area, accumulated change, and both parts of the Fundamental Theorem of Calculus.

From sums to definite integrals

A measures the limiting result of adding many small contributions. Partition $[a,b]$ into $n$ equal subintervals, each of width $\Delta x=\frac{b-a}{n}$ . If $x_i^*$ is a sample point in the $i$ -th subinterval, the corresponding small contribution is $f(x_i^*)\Delta x$ . Adding all contributions produces the

\sum_{i=1}^{n} f(x_i^*)\Delta x.

As the subintervals become narrower, the approximation approaches the :

\int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{i=1}^{n} f(x_i^*)\Delta x,

provided the limit exists. Continuous functions on closed intervals are integrable.

The notation has distinct roles: $a$ and $b$ are the limits of integration, $f(x)$ is the integrand, $x$ is the variable of integration, and $dx$ indicates partitioning with respect to $x$ . The variable is a placeholder, so $\int_a^b f(x)\,dx=\int_a^b f(t)\,dt$ . A is a number, whereas an indefinite integral represents a family of antiderivatives.

Takeaway: Integration begins as a limit of sums and produces a single accumulated value over an interval.

Signed area and integral properties

When $f(x)\ge 0$ throughout $[a,b]$ , the $\int_a^b f(x)\,dx$ equals the ordinary area between the graph and the horizontal axis. When the graph lies below the axis, the contribution is negative. Therefore, the integral represents signed area:

\text{net signed area}=\text{area above the axis}-\text{area below the axis}.

For example, areas of $7$ above the axis and $3$ below the axis give an integral of $7-3=4$ , not a total geometric area of $10$ . To find total geometric area, integrate the absolute value:

\text{total geometric area}=\int_a^b |f(x)|\,dx.

Useful properties include

\int_a^a f(x)\,dx=0, \qquad \int_b^a f(x)\,dx=-\int_a^b f(x)\,dx,

\int_a^b [f(x)+g(x)]\,dx=\int_a^b f(x)\,dx+\int_a^b g(x)\,dx,

\int_a^b c f(x)\,dx=c\int_a^b f(x)\,dx,

and, for any point $c$ ,

\int_a^b f(x)\,dx=\int_a^c f(x)\,dx+\int_c^b f(x)\,dx.

The last property permits an interval to be split when the function changes sign or when separate pieces are easier to interpret.

Takeaway: A records signed area, so determine whether the problem asks for net signed area or total geometric area.

Accumulated change and applications

The connects integration with rates of change. If $r(t)$ is the rate at which a quantity $Q(t)$ changes, then the accumulated from $a$ to $b$ is

Q(b)-Q(a)=\int_a^b r(t)\,dt.

For velocity $v(t)$ , this becomes displacement:

\Delta s=\int_a^b v(t)\,dt.

Because velocity can be negative, displacement may be zero even when the object has traveled a positive distance. For $v(t)=t-2$ on $[0,4]$ ,

\int_0^4 (t-2)\,dt=0,

so the net displacement is zero. However, the velocity is negative on $[0,2]$ and positive on $[2,4]$ . The total distance is

\int_0^4 |t-2|\,dt=4.

More generally, uses the signed rate:

\text{net change}=\int_a^b r(t)\,dt,

whereas uses the magnitude of the rate:

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

To find , locate where $r(t)=0$ , determine the sign on each resulting interval, and integrate the nonnegative pieces.

Takeaway: Integrating a rate gives ; use an absolute value or sign-based intervals when the total amount of movement is required.

The first Fundamental Theorem: differentiation reverses accumulation

The describes differentiation of an accumulation function. Define

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

If $f$ is continuous, then

F'(x)=f(x).

Thus, the derivative of the amount accumulated from $a$ to $x$ is the current value of the integrand. For example, if

F(x)=\int_1^x \sqrt{1+t^2}\,dt,

then

F'(x)=\sqrt{1+x^2}.

No explicit antiderivative of $\sqrt{1+x^2}$ is needed.

If the upper limit is a function of $x$ , apply the chain rule. For

G(x)=\int_2^{x^3}\cos(t^2)\,dt,

first evaluate the integrand at the upper limit and then multiply by the derivative of that limit:

G'(x)=\cos\big((x^3)^2\big)(3x^2)=3x^2\cos(x^6).

Takeaway: For an integral with a variable upper limit, use the integrand at that limit and include the chain-rule factor.

The second Fundamental Theorem: evaluating integrals

The evaluates a through any antiderivative. If $F'(x)=f(x)$ on $[a,b]$ , then

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

The notation

\left[F(x)\right]_a^b

means $F(b)-F(a)$ , or upper endpoint value minus lower endpoint value.

For example, an antiderivative of $3x^2-4x+1$ is $F(x)=x^3-2x^2+x$ , so

\begin{aligned} \int_1^3 (3x^2-4x+1)\,dx &=\left[x^3-2x^2+x\right]_1^3\\ &=(27-18+3)-(1-2+1)\\ &=12. \end{aligned}

For a trigonometric example,

\int_0^{\pi/2}\cos x\,dx=\left[\sin x\right]_0^{\pi/2}=1.

A reliable evaluation procedure is:

Identify the integrand and its limits.
Find an antiderivative.
Substitute the upper endpoint.
Subtract the value at the lower endpoint.
Check the sign, units, and contextual meaning.

Takeaway: The second theorem replaces a limiting sum with the endpoint difference of an antiderivative.

A method for solving and interpreting definite integrals

A practical interpretation should distinguish signed accumulation from total accumulation. For the rate

r(t)=6-2t \quad \text{on } [0,5],

\int_0^5(6-2t)\,dt=5.

The rate changes sign when

6-2t=0,

which gives $t=3$ . Since the rate is positive before $t=3$ and negative afterward, is found by reversing the sign on the negative interval:

\begin{aligned} \int_0^5|6-2t|\,dt &=\int_0^3(6-2t)\,dt+\int_3^5(2t-6)\,dt\\ &=9+4\\ &=13. \end{aligned}

The result means that the quantity has a net gain of $5$ units, while $13$ units moved through the system in both directions. Always check the units: the units of an integral are the units of the integrand multiplied by the units of the variable of integration. A negative answer can be meaningful, representing a decrease, backward displacement, or net loss.

Final checklist:

Identify what is accumulating and over which interval.
Decide whether the question asks for signed or total accumulation.
Use an antiderivative when evaluating a .
Apply upper minus lower carefully.
Split the interval at zeros of the rate when absolute values are required.
Interpret the sign and units in context.