When a rate of change is known, the integral of the rate tells how much the quantity changes. Most applied questions about integrals come down to this one fact, read with the right units.
Net change
If
Q ′ Q' Q ′ is continuous on
[ a , b ] [a, b] [ a , b ] , the integral of the rate of change of
Q Q Q over
[ a , b ] [a, b] [ a , b ] is the net change in
Q Q Q :
∫ a b Q ′ ( t ) d t = Q ( b ) − Q ( a ) . \int_{a}^{b} Q'(t)\,dt = Q(b) - Q(a). ∫ a b Q ′ ( t ) d t = Q ( b ) − Q ( a ) . So the amount at time
t t t is the starting amount plus the change since then:
Q ( t ) = Q ( a ) + ∫ a t Q ′ ( s ) d s . Q(t) = Q(a) + \int_{a}^{t} Q'(s)\,ds. Q ( t ) = Q ( a ) + ∫ a t Q ′ ( s ) d s . The units of the integral are the units of the rate times the units of the variable. A rate in liters per minute, integrated over minutes, gives liters. An interpretation names the quantity, its units, and the interval.
Example 1
A 3D printer draws plastic filament off a spool at a rate of
F ( t ) F(t) F ( t ) grams per minute,
t t t minutes after a print starts, for
0 ≤ t ≤ 40 0 \le t \le 40 0 ≤ t ≤ 40 . The spool holds 750 grams at
t = 0 t = 0 t = 0 . (a) Explain the meaning of the equation below in context.
∫ 5 20 F ( t ) d t = 63 \int_{5}^{20} F(t)\,dt = 63 ∫ 5 20 F ( t ) d t = 63 (b) Write an expression for the amount of filament on the spool at
t = 30 t = 30 t = 30 . (c) Explain the meaning of the equation below in context.
∫ 0 30 F ′ ( t ) d t = − 1.5 \int_{0}^{30} F'(t)\,dt = -1.5 ∫ 0 30 F ′ ( t ) d t = − 1.5 (a) From
t = 5 t = 5 t = 5 to
t = 20 t = 20 t = 20 minutes, the printer draws 63 grams of filament off the spool.
(b) The spool loses filament at the rate
F ( t ) F(t) F ( t ) , so the amount at
t = 30 t = 30 t = 30 , in grams, is
750 − ∫ 0 30 F ( t ) d t . 750 - \int_{0}^{30} F(t)\,dt. 750 − ∫ 0 30 F ( t ) d t . (c) The integral of
F ′ F' F ′ is
F ( 30 ) − F ( 0 ) F(30) - F(0) F ( 30 ) − F ( 0 ) , a change in the rate itself. The rate at which the printer draws filament is 1.5 grams per minute lower at
t = 30 t = 30 t = 30 than at
t = 0 t = 0 t = 0 .
Rates in and rates out When a quantity gains at a rate
I ( t ) I(t) I ( t ) and loses at a rate
O ( t ) O(t) O ( t ) , its rate of change is
I ( t ) − O ( t ) I(t) - O(t) I ( t ) − O ( t ) . The amount is
A ( t ) = A ( a ) + ∫ a t ( I ( s ) − O ( s ) ) d s . A(t) = A(a) + \int_{a}^{t} \big(I(s) - O(s)\big)\,ds. A ( t ) = A ( a ) + ∫ a t ( I ( s ) − O ( s ) ) d s . When
I I I and
O O O are continuous,
A ′ ( t ) = I ( t ) − O ( t ) A'(t) = I(t) - O(t) A ′ ( t ) = I ( t ) − O ( t ) by the Fundamental Theorem. The largest and smallest amounts on a closed interval occur at an end or where
I − O I - O I − O changes sign.
Example 2
A calculator is allowed, and answers are to three decimal places. Bikes are returned to a bike-share dock at a rate of
R ( t ) = 1 + 4 ln ( 1 + t ) R(t) = 1 + 4\ln(1 + t) R ( t ) = 1 + 4 ln ( 1 + t ) bikes per hour. They're taken from it at a rate of
T ( t ) = 9 − 0.6 t T(t) = 9 - 0.6t T ( t ) = 9 − 0.6 t bikes per hour. Here
t t t is hours after 7:00 AM, for
0 ≤ t ≤ 10 0 \le t \le 10 0 ≤ t ≤ 10 , and the dock holds 20 bikes at 7:00 AM.
(a) Find the value of the integral below, and explain its meaning in context.
∫ 4 8 ( R ( t ) − T ( t ) ) d t \int_{4}^{8} \big(R(t) - T(t)\big)\,dt ∫ 4 8 ( R ( t ) − T ( t ) ) d t (b) Find the rate at which the number of bikes at the dock is changing at 1:00 PM, with units. (c) Find the least number of bikes at the dock from 7:00 AM to 5:00 PM, and justify it.
Bikes returned, R, and bikes taken, T. They cross where t is about 3.422. Let
N ( t ) N(t) N ( t ) be the number of bikes at the dock. Then
N ( t ) = 20 + ∫ 0 t ( R ( s ) − T ( s ) ) d s , N(t) = 20 + \int_{0}^{t} \big(R(s) - T(s)\big)\,ds, N ( t ) = 20 + ∫ 0 t ( R ( s ) − T ( s ) ) d s , and since
R R R and
T T T are continuous,
N ′ ( t ) = R ( t ) − T ( t ) N'(t) = R(t) - T(t) N ′ ( t ) = R ( t ) − T ( t ) .
∫ 4 8 ( R ( t ) − T ( t ) ) d t ≈ 13.311. \int_{4}^{8} \big(R(t) - T(t)\big)\,dt \approx 13.311. ∫ 4 8 ( R ( t ) − T ( t ) ) d t ≈ 13.311. It's the net change in the number of bikes at the dock from 11:00 AM to 3:00 PM. The dock gains about 13.311 bikes over those four hours.
(b) At 1:00 PM,
t = 6 t = 6 t = 6 :
N ′ ( 6 ) = R ( 6 ) − T ( 6 ) ≈ 8.784 − 5.4 = 3.384. \begin{aligned}
N'(6) &= R(6) - T(6) \\
&\approx 8.784 - 5.4 = 3.384.
\end{aligned} N ′ ( 6 ) = R ( 6 ) − T ( 6 ) ≈ 8.784 − 5.4 = 3.384. The number of bikes is increasing at about 3.384 bikes per hour.
(c) R R R is increasing and
T T T is decreasing, so
R − T R - T R − T is increasing and has at most one zero. A calculator gives it at
t = c ≈ 3.422 t = c \approx 3.422 t = c ≈ 3.422 , where
N ′ N' N ′ changes from negative to positive. Check that point and the ends:
N ( 0 ) = 20 , N ( c ) = 20 + ∫ 0 c ( R ( t ) − T ( t ) ) d t ≈ 8.744 , N ( 10 ) = 20 + ∫ 0 10 ( R ( t ) − T ( t ) ) d t ≈ 35.507. \begin{aligned}
N(0) &= 20, \\[4pt]
N(c) &= 20 + \int_{0}^{c} \big(R(t) - T(t)\big)\,dt \\
&\approx 8.744, \\[4pt]
N(10) &= 20 + \int_{0}^{10} \big(R(t) - T(t)\big)\,dt \\
&\approx 35.507.
\end{aligned} N ( 0 ) N ( c ) N ( 10 ) = 20 , = 20 + ∫ 0 c ( R ( t ) − T ( t ) ) d t ≈ 8.744 , = 20 + ∫ 0 10 ( R ( t ) − T ( t ) ) d t ≈ 35.507. The least number is about 8.744 bikes, at about 10:25 AM.
Rates with respect to distance The variable doesn't have to be time. A rate per meter, per kilometer, or per unit of any input integrates the same way. The integral's units are the rate's units times the input's units.
Example 3
A cycling route is 6 kilometers long. At a point
x x x kilometers from the start, the elevation changes at a rate of
E ′ ( x ) = 15 ( x − 1 ) ( 5 − x ) E'(x) = 15(x - 1)(5 - x) E ′ ( x ) = 15 ( x − 1 ) ( 5 − x ) meters per kilometer. The start is at an elevation of 350 meters. (a) Find the elevation at the finish. (b) Find the total climbing and the total descent over the route.
Expanding,
E ′ ( x ) = 15 ( − x 2 + 6 x − 5 ) E'(x) = 15\left(-x^2 + 6x - 5\right) E ′ ( x ) = 15 ( − x 2 + 6 x − 5 ) , with antiderivative
G ( x ) = 15 ( − x 3 3 + 3 x 2 − 5 x ) . G(x) = 15\left(-\frac{x^3}{3} + 3x^2 - 5x\right). G ( x ) = 15 ( − 3 x 3 + 3 x 2 − 5 x ) . Its values at
0 0 0 ,
1 1 1 ,
5 5 5 , and
6 6 6 are
0 0 0 ,
− 35 -35 − 35 ,
125 125 125 , and
90 90 90 .
(a) The elevation at the finish, in meters, is
E ( 6 ) = 350 + ∫ 0 6 E ′ ( x ) d x = 350 + ( 90 − 0 ) = 440. \begin{aligned}
E(6) &= 350 + \int_{0}^{6} E'(x)\,dx \\
&= 350 + (90 - 0) = 440.
\end{aligned} E ( 6 ) = 350 + ∫ 0 6 E ′ ( x ) d x = 350 + ( 90 − 0 ) = 440. (b) The rate is negative on
( 0 , 1 ) (0, 1) ( 0 , 1 ) , positive on
( 1 , 5 ) (1, 5) ( 1 , 5 ) , and negative on
( 5 , 6 ) (5, 6) ( 5 , 6 ) . The climbing is the integral over the piece where the route rises:
∫ 1 5 E ′ ( x ) d x = 125 − ( − 35 ) = 160 meters . \begin{aligned}
\int_{1}^{5} E'(x)\,dx &= 125 - (-35) \\
&= 160 \text{ meters}.
\end{aligned} ∫ 1 5 E ′ ( x ) d x = 125 − ( − 35 ) = 160 meters . The descent adds the absolute values of the other two pieces:
∣ ∫ 0 1 E ′ ( x ) d x ∣ + ∣ ∫ 5 6 E ′ ( x ) d x ∣ = ∣ − 35 ∣ + ∣ 90 − 125 ∣ = 70 meters . \begin{aligned}
&\left|\int_{0}^{1} E'(x)\,dx\right| + \left|\int_{5}^{6} E'(x)\,dx\right| \\
&\qquad = |-35| + |90 - 125| \\
&\qquad = 70 \text{ meters}.
\end{aligned} ∫ 0 1 E ′ ( x ) d x + ∫ 5 6 E ′ ( x ) d x = ∣ − 35∣ + ∣90 − 125∣ = 70 meters .
The rate of change of elevation. The climb on (1, 5) has area 160, and each descent has area 35.