# Math Insight

### Applications of integration

Math 1241, Fall 2019
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Due date: Nov. 15, 2019, 11:59 p.m.
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Total points: 3
1. Definite integrals can be used to compute areas in certain situations.
1. 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
, and their sum is
. As $\Delta t$ gets
, 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 $\displaystyle \int_a^b f(x) \, dx$.

2. 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? $\displaystyle \int_{ ＿ }^{ ＿ }$
dx. (Online, to specify limits of integration, enter them here. Lower limit:
. Upper limit:
.)
Now evaluate the definite integral:
3. 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? $\displaystyle \int_{ ＿ }^{ ＿ }$
dx. (Online, to specify limits of integration, enter them here. Lower limit:
. Upper limit:
.)
Now evaluate the definite integral:

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?
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 $-1$. If we multiply the final answer by $-1$, we get the area
. That works because $x^{2} - 4$ is negative over the entire interval.

4. 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
them together.

Feedback from applet
endpoints:
function:

Write down the integral that represents the area below the axis, and evaluate it:
$\displaystyle \int_{ ＿ }^{ ＿ }$
$dx =$

(Online, enter integral limits here. Lower limit:
. Upper limit:
.)

Write down the integral that represents the area above the axis, and evaluate it:
$\displaystyle \int_{ ＿ }^{ ＿ }$
$dx =$

(Online, enter integral limits here. Lower limit:
. Upper limit:
.)

What is the total area bounded by the $x$-axis and $y=x^{2} - 2 x$ between $x=0$ and $x=3$?

5. Evaluate a definite integral (or sum of definite integrals) that gives the area between the $x$-axis and the curve $y=2 e^{2 x}$ between $x=0$ and $x=2$.

In the interval $x \in (0, 2)$, the expression $2 e^{2 x}$ is
. Therefore, we need to split the interval into
pieces, and we
need to change the sign. The area is
.

6. Evaluate a definite integral (or sum of definite integrals) that gives the area between the $x$-axis and the curve $y=- x^{2} + 4 x$ between $x=-1$ and $x=5$.

In the interval $x \in (-1, 5)$, the expression $- x^{2} + 4 x$ is
. We need to split the interval into
pieces.

The intervals are
. Enter in order, separated by commas, such as (-5,2),(2,7).

The sign of $- x^{2} + 4 x$ in the intervals is
. Enter either positive or negative for each interval, in same order as the intervals, separated by commas, such as positive, negative.

. Enter in order, separated by commas, such as (-5,2),(2,7).

The area over or under each interval is
. Enter in order, separated by commas, such 5,3/5.

The total area is
.

2. We can use definite integrals to help us find averages.
1. 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
. Since the total change is given by
and the time that has passed is given by
, we can write down a formula for the average rate of change:
$1$

$\displaystyle \int_{ ＿ }^{ ＿ }$
$dt$.
(Online, enter integral limits here. Lower limit:
. Upper limit:
.)

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:
$1$

$\displaystyle \int_{ ＿ }^{ ＿ }$
$dt$.
(Online, enter integral limits here. Lower limit:
. Upper limit:
.)

2. Find the average value of the function $g(t)=e^{2 t}$ between $t= 0$ and $t=3$. First write down the expression involving an integral that will give the average value, then evaluate it.
$1$

$\displaystyle \int_{ ＿ }^{ ＿ }$
$dt$ $=$

(Online, enter integral limits here. Lower limit:
. Upper limit:
.)

3. A disease is infecting people at a rate of $10 t^{2} + 20 t + 10$ people per day, where $t$ gives time in days. Write down an expression involving an integral that gives the average rate of infection over the first two weeks and evaluate it.
$1$

$\displaystyle \int_{ ＿ }^{ ＿ }$
$dt$ $=$

(Online, enter integral limits here. Lower limit:
. Upper limit:
.)