Introduction
f(b) - f(a)
is the accumulated change of
f
on
[a,b]
.
Therefore, the integral of a derivative
f'
on
[a,b]
is the accumulated change of
f
on
[a,b]
.
f(b) - f(a) = \int_{a}^{b} f'(t) \, dt
The equation above cancels out accumulated change of different signs. If the function increases by 5 units and decreases by 3, the net change is 2 units.
For the total change of a function, which sums all changes regardless of sign, we use:
\int_{a}^{b} |f'(t)| \, dt
Calculating the integral of a function's absolute value involves the following steps:
Find each point where the function changes sign on the domain.
Separate the integral into intervals so that the function does not change sign within any individual integral.
Calculate the integrals.
Sum the absolute values of the integrals.Examples
Example 1: Rocket Motion Analysis
A rocket's velocity is
v(t) = 3t^2 - 12t + 9 \, {m/s}
on
[0, 4]
.
Find
The accumulated change of the rocket's position from t = 0
to t = 4
.
The total distance travelled by the rocket.
\begin{align*}
\int_{0}^{4} (3t^2 - 12t + 9) \, dt &= \left[ t^3 - 6t^2 + 9t \right]_0^4 \\
&= (64 - 96 + 36) - 0 \\
&= 4 \, \text{m} \\
\end{align*}
The rocket ended 4m from its start point.
\text{When } [t = 1 \text{ or } t=3] \Rightarrow [v(t) = 0] \, \\ \text{} \\
\begin{align*}
\int_{0}^{4} |3t^2 - 12t + 9| \, dt &= |\int_{0}^{1} v(t) \, dt| + |\int_{1}^{3} v(t) \, dt| + |\int_{3}^{4} v(t) \, dt| \\ \text{} \\
&= |4| + |-4| + |4| = 12 \, \text{m}
\end{align*}
While the rocket ended
4m
from its start point, it travelled
12m
total.
Example 2: Bacterial Population
A bacteria colony is growing at
P'(t) = 200e^{0.2t}
bacteria/day. Find the population increase from
day 5 to day 10.
\begin{align*}
\Delta P &= \int_{5}^{10} 200e^{0.2t} \, dt \\
&= \left[ 1000e^{0.2t} \right]_5^{10} \\
&= 1000(e^{2} - e^{1}) \\
&= 1000(7.389 - 2.718) \\
&= 4671 \, \text{bacteria}
\end{align*}
There were 4671 more bacteria on day 10 than on day 5.
Example 3: Cost of Production
A factory's costs increase as they produce more units at the rate of
C'(x) = 50\sqrt{x} + 100
dollars/unit. Find the cost
increase from producing 16 to 25 units.
\begin{align*}
\Delta C &= \int_{16}^{25} (50x^{1/2} + 100) \, dx \\
&= \left[ \frac{100}{3}x^{3/2} + 100x \right]_{16}^{25} \\
&= \left(\frac{100}{3}(125) + 2500\right) - \left(\frac{100}{3}(64) + 1600\right) \\
&= \left(4166.67 + 2500\right) - \left(2133.33 + 1600\right) \approx \$2933.34
\end{align*}
The cost of production increased by $2933.34 when the factory increased its production from 16 to 25 units.