The Definite Integral

A definite integral is a limit of Riemann sums, and Riemann sums can be made from any integrand function $f$, positive or negative, continuous or discontinuous. The definite integral still has a geometric meaning even if the function is sometimes (or always) negative, and definite integrals of negative functions also have interpretations in applications.

Example 5: Find the definite integral of $f(x)=-2$ on the interval $[1,4]$.

Solution: $\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{f}\left(\mathrm{c}_{\mathrm{k}}\right) \Delta x_{\mathrm{k}}=\sum_{\mathrm{k}=1}^{\mathrm{n}}(-2) \cdot \Delta x_{\mathrm{k}}=-2 \cdot \sum_{\mathrm{k}=1}^{\mathrm{n}} \Delta x_{\mathrm{k}}=-2 \cdot(3)=-6$

for every partition $P$ and every choice of values for $\mathrm{c}_{\mathrm{k}}$ so $\int_{1}^{4}-2 \mathrm{~d} x=\lim _{\operatorname{mes} h \rightarrow 0}\left(\sum_{k=1}^{n} f\left(c_{k}\right) \cdot \Delta x_{k}\right)=\lim _{\text {mesh } \rightarrow 0}(-6)=-6$.

The area of the region in Fig. 4 is 6 units, but because the region is below the x–axis, the value of the integral is –6 .

If the graph of $f(x)$ is below the x–axis for $\mathrm{a} \leq \mathrm{x} \leq \mathrm{b}$ $f$ is negative),

then $\int_{b}^{a} f(x) \mathrm{d} x= -\{\text { area below the } x \text {-axis for } a \leq x \leq b\}$, a negative number.

If $\mathrm{f}(t)$ is the rate of population change (people/year) for a town, then negative values of $f$ would indicate that the population of the town was getting smaller, and the definite integral (now a negative number) would be the change in the population, a decrease, during the time interval.

Example 6: In 1980 there were 12,000 ducks nesting around a lake, and the rate of population change is shown in Fig. 5. Write a definite integral to represent the total change in the duck population from 1980 to 1990, and estimate the population in 1990.

Solution: Total change in population

$\begin{aligned} &=\int_{1980}^{1990} \mathrm{f}(t) \mathrm{d} t=-\{\text { area between } \mathrm{f} \text { and axis }\}  \\ &\approx-\{200 \text { ducks } / \text { year }\} \cdot\{10 \text { years }\}=-2000 \text { ducks. } \end{aligned}$.

$\begin{aligned} \text { Then }\{1990 \text { duck population }\} &=\{1980 \text { population }\}+\{\text { change from } 1980 \text { to } 1990\} \\ &=\{12,000\}+\{-2000\}=10,000 \text { ducks. } \end{aligned}.$

If $\mathrm{f}(t)$ is the velocity of a car in the positive direction along a straight line at time $t$ (miles/hour) , then negative values of f indicate that the car is traveling in the negative direction. The definite integral of $f$ (the integral is a negative number) is the change in position of the car during the time interval, how far the car traveled in the negative direction.

Practice 3: A bug starts at the location $x = 12$ on the x–axis at 1 pm walks along the axis with the velocity shown in Fig. 6. How far does the bug travel between 1 pm and 3 pm, and where is the bug at 3 pm?

Frequently our integrand functions will be positive some of the time and negative some of the time. If $f$ represents the rate of population increase, then the integral of the positive parts of $f$ will be the increase in population and the integral of the negative parts of $f$ will be the decrease in population. Altogether, the integral of $f$ over the whole time interval will be the total (net) change in the population.

$\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(x) \mathrm{d} x=\{\text { area above axis }\}-\{\text { area below axis }\}$

Example 7: Use Fig. 7 to calculate $\int_{0}^{2} \mathrm{f}(x) \mathrm{d} x, \int_{2}^{4} \mathrm{f}(x) \mathrm{d} x, \int_{4}^{5} \mathrm{f}(x) \mathrm{d} x, \int_{0}^{5} \mathrm{f}(x) \mathrm{d} x$.

Solution: $\int_{0}^{2} \mathrm{f}(x) \mathrm{d} x=2, \int_{2}^{4} \mathrm{f}(x) \mathrm{d} x=-\mathbf{5}, \quad \int_{4}^{5} \mathrm{f}(x) \mathrm{d} x=2$ and $\int_{0}^{5} \mathrm{f}(x) \mathrm{dx}=\{\text { area above }\}-\{\text { area below }\}=\{2+2\}-\{5\}=-1$.

Practice 4: Use geometric reasoning to evaluate $\int_{0}^{2 \pi} \sin (x) \mathrm{d} x$.

If f is a velocity, then the integrals on the intervals where $f$ is positive measure the distances moved forward; the integrals on the intervals where $f$ is negative measure the distances moved backward; and the integral over the whole time interval is the total (net) change in position, the distance moved forward minus the distance moved backward.

Practice 5: A car is driven with the velocity west shown in Fig. 8. (a) Between noon and 6 pm how far does the car travel? (b) At 6 pm, where is the car relative to its starting point (its position at noon)?