What A Definite Integral Measures

A car’s velocity is not constant. At time $t=0$ it might be $3$ m/s, later it might be $8$ m/s, and it can even dip. If you want the total distance traveled from $t=0$ to $t=5$, adding a single rate times a single time does not fit the situation, because there is no one rate that stays true for the whole trip.

What you actually want is total change built from many small changes. Over a tiny time slice, velocity is almost constant, so distance is approximately velocity times that tiny time. If you add those tiny distances over the whole interval, you get the total distance.

That is the problem a definite integral solves. It turns a varying rate into a total accumulation. The punchline you will use repeatedly in this course is that the definite integral measures net accumulation as area under a curve. Soon you will connect this to the Fundamental Theorem of Calculus, but first the meaning of the integral has to feel concrete.

Net accumulation can be negative

Imagine a rate function that is sometimes positive and sometimes negative. For velocity, negative means moving backward. For a cashflow rate, negative means spending. When the rate is negative, the total change should decrease.

On a graph of $y=f(x)$, positive outputs sit above the $x$-axis and negative outputs sit below it. The definite integral counts area above the axis as positive contribution and area below the axis as negative contribution. That is why it measures net change, not just how much happened ignoring direction.

Definition The definite integral $\int_a^b f(x)\,dx$ is interpreted as the signed area between the graph of $y=f(x)$ and the $x$-axis from $x=a$ to $x=b$.

A common mistake is to treat any area as automatically positive. That intuition comes from geometry class, where area means physical size. In accumulation problems, direction matters. If you move forward $10$ meters and then backward $6$ meters, your net change is $4$ meters, not $16$.

Look at a curve that crosses the $x$-axis and decide whether the net accumulation should be positive, negative, or zero based on which signed area dominates.

If the negative region is larger in magnitude than the positive region, the integral is negative, even though the geometric area you could measure with a planimeter would be positive.

Approximating with rectangles

How do you get a number for that signed area when the curve is not a simple shape. The key move is to replace the curve by many thin rectangles whose areas you can add.

Take $f(x)=x^2$ on $[0,2]$. Split the interval into $n$ equal pieces, each of width

$$\Delta x=\frac{2-0}{n}=\frac{2}{n}.$$

On each small subinterval, pick a sample point $x_i^*$, evaluate the function there, and build a rectangle with height $f(x_i^*)$ and width $\Delta x$. The area of that rectangle is approximately

$$f(x_i^*)\Delta x.$$

Add them all to approximate the total area.

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

For $x^2$ on $[0,2]$, the curve is increasing, so left-endpoint rectangles tend to underestimate and right-endpoint rectangles tend to overestimate. As $n$ grows, both estimates squeeze toward the same value, which is what the definite integral means.

Increase the number of rectangles and watch how the estimate settles. What number do the approximations seem to approach.