Math Insight

When we first defined the definite integral, it was as the limit of Riemann sums. The Riemann sum approximated a function as constant on intervals. If we look at this graphically, we can see the Riemann sum from $a$ to $b$ as an approximation of the area under the curve from $x=a$ to $x=b$. Each term in the Riemann sum gives the area of a rectangle with height $f(x_i)$ and width $\Delta t$. Think about how the area of these rectangles relates to the area under the curve: the rectangles fill up the width of the interval [a,b] are close to the height of the curve all of the above are disjoint
, and their sum is approximately the area under the curve equal to the area under the curve unrelated to the area under the curve
. As $\Delta t$ gets smaller larger
, the rectangles get closer to representing the total area under the curve. This means that we can find the area under a curve $y=f(x)$ between $x=a$ and $x=b$ by evaluating the definite integral $\displaystyle \int_a^b f(x) \, dx$.

Let's do a couple quick examples: find the area under the curve $y=3 x + 2$ between $x=0$ and $x=4$. What is the integral that will give this area? (To enter this online, we can use Integral(f(x),(x,a,b)) to represent $\int_a^b f(x)\, dx$.)


Now evaluate the definite integral:

Find the area between the $x$-axis and the curve $y=x^{2} - 4$ between $x=-2$ and $x=2$. What is the integral that will give this area? (To enter this online, we can use Integral(f(x),(x,a,b)) to represent $\int_a^b f(x)\, dx$.)


Now evaluate the definite integral:

What's wrong with this answer? It's positive. It's negative. Nothing.
If we graph $y=x^{2} - 4$, we see that it is below the axis between $-2$ and $2$. What is $f(x_i)$ in our Riemann sums? Positive. Negative. Zero.
In order for the Riemann sum to represent the area, we have to take $|f(x_i)|$ instead of just $f(x_i)$. The definite integral for the area becomes $\displaystyle \int_a^b |f(x)| \, dx$. In this example, we can avoid the absolute values by just multiplying everything by $-1$, yielding
. That works because $x^{2} - 4$ is negative over the entire interval.

How can we handle area for a curve that goes above and below the axis? Suppose we want to know the area bounded by $y=x^{2} - 2 x$ between $x=0$ and $x=3$. Sketch the graph of this function and the region of integration below. There are two regions: one is below the $x$-axis between $x=$
and $x=$
and the other is above the $x$-axis between $x=$
and $x=$
. If we find the area in each of these two regions, we can find the total area by multiplying subtracting adding
them together.

Write down the integral that represents the area below the axis, and evaluate it:

$=$

Write down the integral that represents the area above the axis, and evaluate it:

$=$

Suppose we know that the volume of a cell is changing at a rate of $f(t)$ $\mu {\rm m}^3/s$ and we want to know the average rate from $t=a$ to $t=b$ seconds. The average rate of change is given by the time that has passed divided by the total change the total change divided by the time that has passed the total change times the time that has passed
. Since the total change is given by the definite integral of f the indefinite integral of f the derivative of f
and the time that has passed is given by
, we can write down a formula for the average rate of change:

Though we were interpreting $f(t)$ as the average rate of change, the formula gives us the average value of $f(t)$ between $t=a$ and $t=b$ for any meaning of the function $f(t)$. Suppose we want to know the average volume of the cell between $t=a$ and $t=b$, where the volume at time $t$ is given by $F(t)$. The formula for this is:

A disease is infecting people at a rate of $10 t^{2} + 20 t + 10$ people per day. Write down an expression involving an integral that gives the average rate of infection over the first two weeks and evaluate it.

$=$

The population $w(t)$ of wolves satisfies the differential equation $\frac{dw}{dt}=10 e^{- \frac{t}{20}}$ wolves per year. If the population of wolves at $t=0$ is $300$, what is the average population over the first $3$ years?

Write down an expression involving a definite integral that will give us the average wolf population over the first $3$ years:

In order to evaluate this, we first need to find $w(t)$ by solving the given differential equation. The solution is $w(t)=$
.

Now plug this in to the expression above and evaluate it to find the average population (if rounding, include at least 5 significant figures):