Indefinite integrals and net change
The Fundamental Theorem of Calculus connects integrals with antiderivatives. So we use the integral sign for antiderivatives too. In this section, we introduce this notation. Then we see what an integral of a rate of change means.
Indefinite integrals
The indefinite integral of $f$ is its general antiderivative:
$\displaystyle\int f(x)\,dx=F(x)+C$
means that $F'(x)=f(x)$.
For example,
$\displaystyle\int x^2\,dx=\dfrac{x^3}{3}+C$,
because the derivative of $\dfrac{x^3}{3}$ is $x^2$.
Notice the difference:
- A definite integral $\displaystyle\int_a^b f(x)\,dx$ has limits. It is a number.
- An indefinite integral $\displaystyle\int f(x)\,dx$ has no limits. It is a family of functions, one for each value of $C$.
They are connected by the Fundamental Theorem: to find a definite integral, find the indefinite integral, then subtract its values at the two ends.
A table of indefinite integrals
These come from the antiderivative formulas. Each can be checked by differentiating the right side.
- $\displaystyle\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C$ ($n\ne -1$)
- $\displaystyle\int\dfrac{1}{x}\,dx=\ln|x|+C$
- $\displaystyle\int e^x\,dx=e^x+C$
- $\displaystyle\int\cos x\,dx=\sin x+C$
- $\displaystyle\int\sin x\,dx=-\cos x+C$
- $\displaystyle\int\sec^2 x\,dx=\tan x+C$
- $\displaystyle\int\sec x\tan x\,dx=\sec x+C$
Also, constants move out, and sums can be integrated one term at a time:
$\displaystyle\int c\,f(x)\,dx=c\int f(x)\,dx$
$\begin{align*}&\int[f(x)\pm g(x)]\,dx\\&=\int f(x)\,dx\pm\int g(x)\,dx\end{align*}$
Example 1: Find
$\displaystyle\int(10x^4-2\sec^2 x)\,dx$.
Solution:
$\begin{align*}&\int(10x^4-2\sec^2 x)\,dx\\&=10\cdot\dfrac{x^5}{5}-2\tan x+C\\&=2x^5-2\tan x+C\end{align*}$
Check: The derivative of $2x^5-2\tan x$ is $10x^4-2\sec^2 x$.
Example 2: Find
$\displaystyle\int\dfrac{x^2+1}{\sqrt{x}}\,dx$.
Solution:
There is no quotient rule for integrals. So simplify first. Divide each term by $x^{1/2}$ (that is, $\sqrt{x}$):
$\dfrac{x^2+1}{\sqrt{x}}=x^{3/2}+x^{-1/2}$
Now integrate one term at a time:
$\begin{align*}&\int(x^{3/2}+x^{-1/2})\,dx\\&=\dfrac{x^{5/2}}{5/2}+\dfrac{x^{1/2}}{1/2}+C\\&=\dfrac{2}{5}x^{5/2}+2x^{1/2}+C\end{align*}$
Example 3: Find
$\displaystyle\int\dfrac{\sin\theta}{\cos^2\theta}\,d\theta$.
Solution:
This is not in the table. But split the fraction:
$\dfrac{\sin\theta}{\cos^2\theta}=\dfrac{1}{\cos\theta}\cdot\dfrac{\sin\theta}{\cos\theta}=\sec\theta\tan\theta$
So
$\displaystyle\int\dfrac{\sin\theta}{\cos^2\theta}\,d\theta=\sec\theta+C$.
Example 4: Find
$\displaystyle\int_1^9\dfrac{2t^2+t^{5/2}-1}{t^2}\,dt$.
Solution:
Divide each term by $t^2$ first:
$\dfrac{2t^2+t^{5/2}-1}{t^2}=2+t^{1/2}-t^{-2}$
An antiderivative is $2t+\dfrac{2}{3}t^{3/2}+\dfrac{1}{t}$. Now use the Fundamental Theorem:
$\begin{align*}&\left[2t+\dfrac{2}{3}t^{3/2}+\dfrac{1}{t}\right]_1^9\\&=\left(18+18+\dfrac{1}{9}\right)-\left(2+\dfrac{2}{3}+1\right)\\&=\dfrac{325}{9}-\dfrac{33}{9}\\&=\dfrac{292}{9}\approx 32.44\end{align*}$
Here $9^{3/2}=(\sqrt{9})^3=27$.
So $\dfrac{2}{3}\cdot 27=18$.
The net change theorem
If $F'$ is the rate of change of $F$, then the Fundamental Theorem says
$\displaystyle\int_a^b F'(x)\,dx=F(b)-F(a)$.
The right side is the change in $F$ from $a$ to $b$. This gives the net change theorem:
The integral of a rate of change is the net change.
Here are some examples:
- If water flows into a tank at a rate $r(t)$, then $\displaystyle\int_a^b r(t)\,dt$ is the amount of water added from time $a$ to time $b$.
- If a population grows at a rate $P'(t)$, then $\displaystyle\int_a^b P'(t)\,dt$ is the change in the population.
- If an object has velocity $v(t)$, then $\displaystyle\int_a^b v(t)\,dt$ is the change in its position.
Example 5: Water flows into a tank at a rate of
$r(t)=200-4t$ liters per minute,
for $0\le t\le 50$. How much water flows in during the first $10$ minutes?
Solution:
By the net change theorem, the amount is the integral of the rate:
$\begin{align*}&\int_0^{10}(200-4t)\,dt\\&=\Big[200t-2t^2\Big]_0^{10}\\&=(2000-200)-0\\&=1800\end{align*}$
So $1800$ liters flow in.
Displacement and distance
For an object moving on a line with velocity $v(t)$:
- The displacement is the change in position: $\displaystyle\int_a^b v(t)\,dt$. Backward motion counts as negative.
- The total distance traveled counts all motion as positive: $\displaystyle\int_a^b|v(t)|\,dt$.
To find the total distance, split the time interval where $v$ changes sign. Integrate each piece, and add the sizes (absolute values) of the results.
Example 6: A particle moves on a line with velocity
$v(t)=t^2-t-6$ m/s.
(a) Find its displacement for $1\le t\le 4$.
(b) Find the total distance it travels in that time.
Solution:
An antiderivative of $v$ is
$F(t)=\dfrac{t^3}{3}-\dfrac{t^2}{2}-6t$.
Its values: $F(1)=-\dfrac{37}{6}$, $F(3)=-\dfrac{27}{2}$, and $F(4)=-\dfrac{32}{3}$.
(a) The displacement:
$\begin{align*}&\int_1^4 v(t)\,dt\\&=F(4)-F(1)\\&=-\dfrac{32}{3}+\dfrac{37}{6}\\&=-\dfrac{9}{2}\end{align*}$
The particle ends $4.5$ m behind where it started.
(b) Factor the velocity:
$v(t)=(t-3)(t+2)$
- On $(1,3)$: $v(t)<0$.
- On $(3,4)$: $v(t)>0$.
Integrate each piece:
$\begin{align*}&\int_1^3 v(t)\,dt\\&=F(3)-F(1)\\&=-\dfrac{27}{2}+\dfrac{37}{6}\\&=-\dfrac{22}{3}\end{align*}$
$\begin{align*}&\int_3^4 v(t)\,dt\\&=F(4)-F(3)\\&=-\dfrac{32}{3}+\dfrac{27}{2}\\&=\dfrac{17}{6}\end{align*}$
Add the sizes:
$\dfrac{22}{3}+\dfrac{17}{6}=\dfrac{61}{6}\approx 10.17$ m
The particle moved $\dfrac{22}{3}$ m backward and then $\dfrac{17}{6}$ m forward. Its total distance is about $10.17$ m, but its displacement is only $-4.5$ m.
Summary
- The indefinite integral of $f$ is its general antiderivative, $F(x)+C$. It is a family of functions.
- Simplify first: there is no product or quotient rule for integrals.
- The integral of a rate of change is the net change.
- Displacement is $\displaystyle\int_a^b v(t)\,dt$. Total distance is $\displaystyle\int_a^b|v(t)|\,dt$.
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