You know how to average a list of numbers: add them up and divide by how many there are. But what is the average temperature over a whole day, when the temperature changes every instant? A definite integral does the “adding up” for infinitely many values, and dividing by the length of the interval finishes the job. That’s the average value of a function .
The average value of a continuous function f f f [ a , b ] [a, b] [ a , b ]
f avg = 1 b − a ∫ a b f ( x ) d x f_{\text{avg}} = \frac{1}{b - a} \int_a^b f(x)\, dx f avg = b − a 1 ∫ a b f ( x ) d x Compare it with averaging a list: the integral plays the role of the sum, and the length b − a b - a b − a
Multiply both sides by b − a b - a b − a
f avg ⋅ ( b − a ) = ∫ a b f ( x ) d x f_{\text{avg}} \cdot (b - a) = \int_a^b f(x)\, dx f avg ⋅ ( b − a ) = ∫ a b f ( x ) d x So f avg f_{\text{avg}} f avg rectangle on [ a , b ] [a, b] [ a , b ]
The area under y = x squared from 0 to 3 equals the area of a rectangle of height 3 on the same base. The curve meets height 3 at x = root 3.
c = √3
y = x²
average = 3
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For y = x 2 y = x^2 y = x 2 [ 0 , 3 ] [0, 3] [ 0 , 3 ] 9 9 9 9 3 = 3 \dfrac{9}{3} = 3 3 9 = 3 x = 3 x = \sqrt{3} x = 3 If f f f [ a , b ] [a, b] [ a , b ] c c c [ a , b ] [a, b] [ a , b ]
f ( c ) = f avg f(c) = f_{\text{avg}} f ( c ) = f avg This is sometimes called the Mean Value Theorem for integrals . It makes sense from the picture: a continuous curve can’t stay entirely above or entirely below the rectangle, so it has to cross the line y = f avg y = f_{\text{avg}} y = f avg c c c f ( c ) f(c) f ( c )
The average value has the same units as f f f , not the units of the integral. If T ( t ) T(t) T ( t ) t t t ∫ T d t \int T\, dt ∫ T d t
These sound alike but answer different questions:
Formula Meaning Average value of f f f 1 b − a ∫ a b f ( x ) d x \dfrac{1}{b - a}\displaystyle\int_a^b f(x)\, dx b − a 1 ∫ a b f ( x ) d x the typical height of f f f Average rate of change of f f f f ( b ) − f ( a ) b − a \dfrac{f(b) - f(a)}{b - a} b − a f ( b ) − f ( a ) the slope of the secant line
They connect nicely: the average value of f ′ f' f ′ [ a , b ] [a, b] [ a , b ] 1 b − a ∫ a b f ′ ( x ) d x = f ( b ) − f ( a ) b − a \dfrac{1}{b - a}\displaystyle\int_a^b f'(x)\, dx = \dfrac{f(b) - f(a)}{b - a} b − a 1 ∫ a b f ′ ( x ) d x = b − a f ( b ) − f ( a ) f f f
On the AP exam, average value shows up in both calculator and no-calculator questions. When a calculator is allowed, write the setup (the integral with its 1 b − a \frac{1}{b - a} b − a 1
Find the average value of f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 [ 0 , 3 ] [0, 3] [ 0 , 3 ] c c c f ( c ) f(c) f ( c )
Solution.
f avg = 1 3 − 0 ∫ 0 3 x 2 d x = 1 3 [ x 3 3 ] 0 3 = 1 3 ( 9 ) = 3 f_{\text{avg}} = \frac{1}{3 - 0} \int_0^3 x^2\, dx = \frac{1}{3} \left[ \frac{x^3}{3} \right]_0^3 = \frac{1}{3}(9) = 3 f avg = 3 − 0 1 ∫ 0 3 x 2 d x = 3 1 [ 3 x 3 ] 0 3 = 3 1 ( 9 ) = 3 Now solve f ( c ) = 3 f(c) = 3 f ( c ) = 3 c 2 = 3 c^2 = 3 c 2 = 3 c = ± 3 c = \pm\sqrt{3} c = ± 3 c = 3 ≈ 1.732 c = \sqrt{3} \approx 1.732 c = 3 ≈ 1.732 [ 0 , 3 ] [0, 3] [ 0 , 3 ]
Find the average value of f ( x ) = sin x f(x) = \sin x f ( x ) = sin x [ 0 , π ] [0, \pi] [ 0 , π ]
Solution.
f avg = 1 π − 0 ∫ 0 π sin x d x = 1 π [ − cos x ] 0 π = 1 π ( 1 − ( − 1 ) ) = 2 π f_{\text{avg}} = \frac{1}{\pi - 0} \int_0^{\pi} \sin x\, dx = \frac{1}{\pi} \Big[ -\cos x \Big]_0^{\pi} = \frac{1}{\pi} \big( 1 - (-1) \big) = \frac{2}{\pi} f avg = π − 0 1 ∫ 0 π sin x d x = π 1 [ − cos x ] 0 π = π 1 ( 1 − ( − 1 ) ) = π 2 That’s about 0.637 0.637 0.637 1 2 \tfrac{1}{2} 2 1
On a spring day, the temperature t t t T ( t ) = 15 + 6 sin ( π t 12 ) T(t) = 15 + 6\sin\!\left(\dfrac{\pi t}{12}\right) T ( t ) = 15 + 6 sin ( 12 π t )
Solution. The interval is 0 ≤ t ≤ 12 0 \le t \le 12 0 ≤ t ≤ 12
T avg = 1 12 ∫ 0 12 ( 15 + 6 sin π t 12 ) d t = 1 12 [ 15 t − 72 π cos π t 12 ] 0 12 = 1 12 [ ( 180 + 72 π ) − ( 0 − 72 π ) ] = 1 12 ( 180 + 144 π ) = 15 + 12 π \begin{aligned}
T_{\text{avg}} &= \frac{1}{12} \int_0^{12} \left( 15 + 6\sin\frac{\pi t}{12} \right) dt \\
&= \frac{1}{12} \left[ 15t - \frac{72}{\pi} \cos\frac{\pi t}{12} \right]_0^{12} \\
&= \frac{1}{12} \left[ \left( 180 + \frac{72}{\pi} \right) - \left( 0 - \frac{72}{\pi} \right) \right] \\
&= \frac{1}{12} \left( 180 + \frac{144}{\pi} \right) = 15 + \frac{12}{\pi}
\end{aligned} T avg = 12 1 ∫ 0 12 ( 15 + 6 sin 12 π t ) d t = 12 1 [ 15 t − π 72 cos 12 π t ] 0 12 = 12 1 [ ( 180 + π 72 ) − ( 0 − π 72 ) ] = 12 1 ( 180 + π 144 ) = 15 + π 12 The average temperature is 15 + 12 π ≈ 18.820 15 + \dfrac{12}{\pi} \approx 18.820 15 + π 12 ≈ 18.820 T T T
Check: the antiderivative of sin π t 12 \sin\dfrac{\pi t}{12} sin 12 π t − 12 π cos π t 12 -\dfrac{12}{\pi}\cos\dfrac{\pi t}{12} − π 12 cos 12 π t 6 ⋅ 12 π = 72 π 6 \cdot \dfrac{12}{\pi} = \dfrac{72}{\pi} 6 ⋅ π 12 = π 72
The average value of f ( x ) = 6 x 2 f(x) = 6x^2 f ( x ) = 6 x 2 [ 0 , k ] [0, k] [ 0 , k ] 8 8 8 k > 0 k \gt 0 k > 0
Solution.
1 k ∫ 0 k 6 x 2 d x = 1 k [ 2 x 3 ] 0 k = 2 k 3 k = 2 k 2 \frac{1}{k} \int_0^k 6x^2\, dx = \frac{1}{k} \Big[ 2x^3 \Big]_0^k = \frac{2k^3}{k} = 2k^2 k 1 ∫ 0 k 6 x 2 d x = k 1 [ 2 x 3 ] 0 k = k 2 k 3 = 2 k 2 Set 2 k 2 = 8 2k^2 = 8 2 k 2 = 8 k 2 = 4 k^2 = 4 k 2 = 4 k = 2 k = 2 k = 2 k > 0 k \gt 0 k > 0
Forgetting to divide by b − a. The integral alone is the area (or total), not the average. In Example 1, ∫ 0 3 x 2 d x = 9 \int_0^3 x^2\, dx = 9 ∫ 0 3 x 2 d x = 9 3 3 3
Dividing by the wrong length. It’s b − a b - a b − a b b b [ 2 , 6 ] [2, 6] [ 2 , 6 ] 4 4 4 6 6 6
Mixing up average value and average rate of change. “Average value of f f f f f f f f f f ( b ) − f ( a ) b − a \dfrac{f(b) - f(a)}{b - a} b − a f ( b ) − f ( a )
Giving the wrong units. The average value has the units of f f f
Averaging just the endpoints. f ( a ) + f ( b ) 2 \dfrac{f(a) + f(b)}{2} 2 f ( a ) + f ( b ) x 2 x^2 x 2 [ 0 , 3 ] [0, 3] [ 0 , 3 ] 4.5 4.5 4.5 3 3 3
Keeping a c outside the interval. When you solve f ( c ) = f avg f(c) = f_{\text{avg}} f ( c ) = f avg [ a , b ] [a, b] [ a , b ]
1. (Warm-up) Find the average value of f ( x ) = 4 x f(x) = 4x f ( x ) = 4 x [ 0 , 5 ] [0, 5] [ 0 , 5 ]
Solution 1 5 ∫ 0 5 4 x d x = 1 5 [ 2 x 2 ] 0 5 = 1 5 ( 50 ) = 10 \frac{1}{5} \int_0^5 4x\, dx = \frac{1}{5} \Big[ 2x^2 \Big]_0^5 = \frac{1}{5}(50) = 10 5 1 ∫ 0 5 4 x d x = 5 1 [ 2 x 2 ] 0 5 = 5 1 ( 50 ) = 10 Check: f f f 0 + 20 2 = 10 \dfrac{0 + 20}{2} = 10 2 0 + 20 = 10
2. (Warm-up) You are told that ∫ 1 5 g ( x ) d x = 12 \displaystyle\int_1^5 g(x)\, dx = 12 ∫ 1 5 g ( x ) d x = 12 g g g [ 1 , 5 ] [1, 5] [ 1 , 5 ]
Solution g avg = 1 5 − 1 ( 12 ) = 3 g_{\text{avg}} = \frac{1}{5 - 1}(12) = 3 g avg = 5 − 1 1 ( 12 ) = 3 3. (Warm-up) Find the average value of f ( x ) = x 3 f(x) = x^3 f ( x ) = x 3 [ 0 , 2 ] [0, 2] [ 0 , 2 ]
Solution 1 2 ∫ 0 2 x 3 d x = 1 2 [ x 4 4 ] 0 2 = 1 2 ( 4 ) = 2 \frac{1}{2} \int_0^2 x^3\, dx = \frac{1}{2} \left[ \frac{x^4}{4} \right]_0^2 = \frac{1}{2}(4) = 2 2 1 ∫ 0 2 x 3 d x = 2 1 [ 4 x 4 ] 0 2 = 2 1 ( 4 ) = 2 4. (Core) Find the average value of f ( x ) = 1 x f(x) = \dfrac{1}{x} f ( x ) = x 1 [ 1 , e ] [1, e] [ 1 , e ]
Solution 1 e − 1 ∫ 1 e 1 x d x = 1 e − 1 [ ln x ] 1 e = 1 e − 1 ( 1 − 0 ) = 1 e − 1 ≈ 0.582 \frac{1}{e - 1} \int_1^e \frac{1}{x}\, dx = \frac{1}{e - 1} \Big[ \ln x \Big]_1^e = \frac{1}{e - 1}(1 - 0) = \frac{1}{e - 1} \approx 0.582 e − 1 1 ∫ 1 e x 1 d x = e − 1 1 [ ln x ] 1 e = e − 1 1 ( 1 − 0 ) = e − 1 1 ≈ 0.582 5. (Core) Find the average value of f ( x ) = e 2 x f(x) = e^{2x} f ( x ) = e 2 x [ 0 , ln 3 ] [0, \ln 3] [ 0 , ln 3 ]
Solution ∫ 0 ln 3 e 2 x d x = [ e 2 x 2 ] 0 ln 3 = e 2 ln 3 − 1 2 = 9 − 1 2 = 4 \int_0^{\ln 3} e^{2x}\, dx = \left[ \frac{e^{2x}}{2} \right]_0^{\ln 3} = \frac{e^{2\ln 3} - 1}{2} = \frac{9 - 1}{2} = 4 ∫ 0 l n 3 e 2 x d x = [ 2 e 2 x ] 0 l n 3 = 2 e 2 l n 3 − 1 = 2 9 − 1 = 4 (Remember e 2 ln 3 = ( e ln 3 ) 2 = 9 e^{2\ln 3} = \left(e^{\ln 3}\right)^2 = 9 e 2 l n 3 = ( e l n 3 ) 2 = 9
f avg = 4 ln 3 ≈ 3.641 f_{\text{avg}} = \frac{4}{\ln 3} \approx 3.641 f avg = ln 3 4 ≈ 3.641 6. (Core) Find the average value of f ( x ) = sec 2 x f(x) = \sec^2 x f ( x ) = sec 2 x [ 0 , π 4 ] \left[0, \dfrac{\pi}{4}\right] [ 0 , 4 π ]
Solution 1 π / 4 ∫ 0 π / 4 sec 2 x d x = 4 π [ tan x ] 0 π / 4 = 4 π ( 1 − 0 ) = 4 π ≈ 1.273 \frac{1}{\pi/4} \int_0^{\pi/4} \sec^2 x\, dx = \frac{4}{\pi} \Big[ \tan x \Big]_0^{\pi/4} = \frac{4}{\pi}(1 - 0) = \frac{4}{\pi} \approx 1.273 π /4 1 ∫ 0 π /4 sec 2 x d x = π 4 [ tan x ] 0 π /4 = π 4 ( 1 − 0 ) = π 4 ≈ 1.273 7. (Core) A cyclist’s velocity is v ( t ) = t 2 + 2 v(t) = t^2 + 2 v ( t ) = t 2 + 2 0 ≤ t ≤ 4 0 \le t \le 4 0 ≤ t ≤ 4 4 4 4
Solution v avg = 1 4 ∫ 0 4 ( t 2 + 2 ) d t = 1 4 [ t 3 3 + 2 t ] 0 4 = 1 4 ( 64 3 + 8 ) = 1 4 ⋅ 88 3 = 22 3 v_{\text{avg}} = \frac{1}{4} \int_0^4 (t^2 + 2)\, dt = \frac{1}{4} \left[ \frac{t^3}{3} + 2t \right]_0^4 = \frac{1}{4} \left( \frac{64}{3} + 8 \right) = \frac{1}{4} \cdot \frac{88}{3} = \frac{22}{3} v avg = 4 1 ∫ 0 4 ( t 2 + 2 ) d t = 4 1 [ 3 t 3 + 2 t ] 0 4 = 4 1 ( 3 64 + 8 ) = 4 1 ⋅ 3 88 = 3 22 The average velocity is 22 3 ≈ 7.333 \dfrac{22}{3} \approx 7.333 3 22 ≈ 7.333
8. (Challenge) Let f ( x ) = x 2 − 4 x + 5 f(x) = x^2 - 4x + 5 f ( x ) = x 2 − 4 x + 5 f f f [ 0 , 3 ] [0, 3] [ 0 , 3 ] c c c [ 0 , 3 ] [0, 3] [ 0 , 3 ] f ( c ) f(c) f ( c )
Solution ∫ 0 3 ( x 2 − 4 x + 5 ) d x = [ x 3 3 − 2 x 2 + 5 x ] 0 3 = 9 − 18 + 15 = 6 \int_0^3 (x^2 - 4x + 5)\, dx = \left[ \frac{x^3}{3} - 2x^2 + 5x \right]_0^3 = 9 - 18 + 15 = 6 ∫ 0 3 ( x 2 − 4 x + 5 ) d x = [ 3 x 3 − 2 x 2 + 5 x ] 0 3 = 9 − 18 + 15 = 6 So f avg = 6 3 = 2 f_{\text{avg}} = \dfrac{6}{3} = 2 f avg = 3 6 = 2
Solve c 2 − 4 c + 5 = 2 c^2 - 4c + 5 = 2 c 2 − 4 c + 5 = 2 c 2 − 4 c + 3 = 0 c^2 - 4c + 3 = 0 c 2 − 4 c + 3 = 0 ( c − 1 ) ( c − 3 ) = 0 (c - 1)(c - 3) = 0 ( c − 1 ) ( c − 3 ) = 0 c = 1 c = 1 c = 1 c = 3 c = 3 c = 3 [ 0 , 3 ] [0, 3] [ 0 , 3 ]
9. (Challenge) The average value of a continuous function f f f [ 0 , 6 ] [0, 6] [ 0 , 6 ] 5 5 5 [ 0 , 2 ] [0, 2] [ 0 , 2 ] 8 8 8 f f f [ 2 , 6 ] [2, 6] [ 2 , 6 ]
Solution Turn each average back into an integral by multiplying by the length:
∫ 0 6 f ( x ) d x = 6 ( 5 ) = 30 , ∫ 0 2 f ( x ) d x = 2 ( 8 ) = 16 \int_0^6 f(x)\, dx = 6(5) = 30, \qquad \int_0^2 f(x)\, dx = 2(8) = 16 ∫ 0 6 f ( x ) d x = 6 ( 5 ) = 30 , ∫ 0 2 f ( x ) d x = 2 ( 8 ) = 16 So ∫ 2 6 f ( x ) d x = 30 − 16 = 14 \displaystyle\int_2^6 f(x)\, dx = 30 - 16 = 14 ∫ 2 6 f ( x ) d x = 30 − 16 = 14 [ 2 , 6 ] [2, 6] [ 2 , 6 ] 14 6 − 2 = 3.5 \dfrac{14}{6 - 2} = 3.5 6 − 2 14 = 3.5
Notice you can’t just “subtract the averages”: the intervals have different lengths.
