Integration and Accumulation

Read a definite integral as the accumulation of many small contributions. Connect area, total change, antiderivatives, numerical sums, and the probability carried by a continuous density.

An integral accumulates contributions across an interval. It can recover total change from a changing rate, measure signed area, assign probability across a continuous range, and define an average value. The definite integral is a limit of finite sums, while the Fundamental Theorem of Calculus connects that total with antiderivatives.

Accumulation Begins with Small Contributions

Suppose $r(t)$ is a rate measured in units per second. During a short interval of width $\Delta t$ , the accumulated change is approximately:

r(t)\Delta t.

Divide the interval $[a,b]$ into smaller pieces. If the $i$ th piece has width $\Delta t_i$ and a representative point $t_i^*$ , the sum:

\sum_{i=1}^{n}r(t_i^*)\Delta t_i

approximates the total change. Such a sum is called a Riemann sum. As the widest piece shrinks toward zero, the limiting total is the definite integral:

\int_a^b r(t)\,dt.

Read the integral as: “add the contributions $r(t)$ times very small widths $dt$ from $t=a$ to $t=b$ .” It is the limit of increasingly fine finite sums, not ordinary multiplication by the symbol $dt$ .

Read this as “the integral of $r(t)$ from $a$ to $b$ with respect to $t$ .” The symbol $dt$ names the variable being partitioned into small widths.

Units provide a useful check. Multiplying units per second by seconds leaves units of accumulated quantity. An integral of velocity over time therefore has units of distance.

Exercise: Accumulate a constant rate

A process changes at a constant rate of $4$ units per second for $2.5$ seconds. What total change does the integral give?

Check the total change

Compute it first, then check your number.

HintUse a rectangular contribution

A constant height of $4$ across a width of $2.5$ has total $4\times2.5$ .

SolutionMultiply rate by elapsed time

The accumulated change is:

\int_0^{2.5}4\,dt =4(2.5-0) =10.

The seconds in the interval width cancel the per-second part of the rate, leaving $10$ units of total change.

A Definite Integral Is Signed Area

When $f(x)\ge0$ , the integral $\int_a^b f(x)\,dx$ equals the geometric area between the graph and the horizontal axis. Contributions below the axis are negative, so a definite integral measures signed area, not necessarily the total geometric area.

For example:

\int_{-1}^{1}x\,dx=0.

The negative triangular contribution on $[-1,0]$ cancels the positive triangular contribution on $[0,1]$ . The geometric area is $1$ , but the signed integral is $0$ .

Reversing the bounds reverses the sign:

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

and splitting an interval preserves the total:

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

Exercise: Distinguish signed area from geometric area

What is $\int_{-2}^{2}x\,dx$ ?

Check the signed integral

Compute it first, then check your number.

HintUse symmetry

For every positive contribution at $x$ , there is an equal negative contribution at $-x$ .

SolutionCancel equal contributions with opposite signs

The line $y=x$ is below the axis on $[-2,0]$ and above it on $[0,2]$ . The two triangles have equal geometric area and opposite signs. Therefore:

\int_{-2}^{2}x\,dx=0.

The total geometric area would be $4$ , so it is important to identify which quantity the question requests.

Antiderivatives Evaluate Exact Totals

An antiderivative of $f$ is a function $F$ satisfying $F'(x)=f(x)$ . Since adding a constant does not change a derivative, an indefinite integral denotes a family of antiderivatives:

\int f(x)\,dx=F(x)+C.

The Fundamental Theorem of Calculus connects differentiation and accumulation. If $F'=f$ , then:

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

For $f(x)=2x$ , one antiderivative is $F(x)=x^2$ . Hence:

\int_1^3 2x\,dx =3^2-1^2 =8.

The theorem also works in the other direction. If:

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

then, under the usual continuity conditions, $A'(x)=f(x)$ . The local rate of the accumulated total is the contribution being accumulated.

Exercise: Use an antiderivative

Compute $\int_0^2 3x^2\,dx$ using the antiderivative $F(x)=x^3$ .

Check the definite integral

Compute it first, then check your number.

HintUse upper endpoint minus lower endpoint

Apply $F(2)-F(0)$ with $F(x)=x^3$ .

SolutionEvaluate the antiderivative at the bounds

Since $F'(x)=3x^2$ , the Fundamental Theorem gives:

\int_0^2 3x^2\,dx =F(2)-F(0) =2^3-0^3 =8.

No arbitrary constant appears in the final answer because it cancels between the two endpoint evaluations.

Integrals Define Averages and Continuous Probability

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

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

This is total accumulated output divided by interval width. It extends the ordinary arithmetic mean to a continuous range.

A probability density $p(x)$ describes probability per unit of $x$ . Its total area must be one:

\int_{-\infty}^{\infty}p(x)\,dx=1,

and interval probability is accumulated density:

P(a\le X\le b)=\int_a^b p(x)\,dx.

Density height is not itself probability. A density can exceed $1$ on a narrow region as long as its total integral is $1$ . For a continuous random variable, an isolated point has zero interval width and usually has probability zero, even when the density at that point is positive.

Exercise: Normalize a constant density

A probability density is constant on $[0,4]$ and zero elsewhere. What constant height makes its total probability equal to $1$ ?

Check the density height

Compute it first, then check your number.

HintNormalize the total area

If the height is $c$ , then $\int_0^4c\,dx=4c$ .

SolutionSet accumulated density equal to one

Normalization requires:

\int_0^4c\,dx=4c=1.

Therefore:

c=\frac14=0.25.

The units of density are inverse units of $x$ , allowing density times interval width to be a dimensionless probability.

Numerical Integration Approximates the Limiting Sum

Some antiderivatives are difficult or impossible to express with elementary functions. Numerical methods approximate the integral with finitely many function evaluations.

A left-rectangle rule uses the value at the left endpoint of each subinterval. A trapezoidal rule averages the two endpoint heights. On one interval $[a,b]$ , the trapezoidal estimate is:

\int_a^b f(x)\,dx \approx(b-a)\frac{f(a)+f(b)}{2}.

Dividing the interval into more pieces often improves the estimate for a smooth function, but accuracy depends on the function and method. A smaller step also increases computation, and extremely small steps can interact with floating-point rounding.

Exercise: Apply one trapezoidal step

On an interval of width $2$ , a function has endpoint values $1$ and $3$ . What integral estimate does one trapezoid give?

Check the estimate

Compute it first, then check your number.

HintFind the trapezoid's average height

Compute $2(1+3)/2$ .

SolutionMultiply width by average endpoint height

The trapezoidal estimate is:

2\left(\frac{1+3}{2}\right)=2(2)=4.

This estimate is exact when the function is linear across the interval. For a curved function, subdividing the interval can reduce the approximation error.

Refine a trapezoidal approximation

Approximate the integral of x² from 0 to 1. Increase the number of subintervals and compare the estimate with the exact value 1/3.

Ready to run.

Before Interpreting an Integral

Identify what is being accumulated, the variable and interval of accumulation, and the resulting units. Distinguish signed integral from geometric area. When using an antiderivative, evaluate upper bound minus lower bound. When using a numerical method, state the partition or step size and treat the result as an approximation with method-dependent error.

The next lesson returns to local change for functions with several inputs. Partial derivatives isolate the sensitivity to one input while holding the others fixed.