Average value of a function: Worked examples — Maths deep dive

The scenario matrix

Every "average value" problem falls into one of these cells. We will hit all of them.

#	Case class	What's tricky about it	Example that covers it
C1	$f \geq 0$ everywhere	the baseline, area is honest area	Ex 1
C2	$f \leq 0$ everywhere	integral is negative → average negative	Ex 2
C3	$f$ changes sign, symmetric	positive & negative bits cancel → average $= 0$	Ex 3
C4	$f$ changes sign, not symmetric	partial cancellation → small nonzero average	Ex 4
C5	interval in negative $x$	limits like $[- 2, - 1]$ , watch the width	Ex 5
C6	degenerate: $b - a \to 0$	formula becomes $0/0$ → limiting value	Ex 6
C7	real-world word problem	you must find $a, b, f$ yourself	Ex 7
C8	exam twist: find the $c$ (MVT)	solve $f (c) = f_{avg}$ , keep the root in range	Ex 8

The last cell uses the Mean Value Theorem for Integrals (abbreviated MVT from here on — it is the theorem that guarantees a special interior point $c$ where the function equals its own average).

Before we start, one thing to burn in: an integral is a signed area. Bits of the curve above the $x$ -axis count as positive area; bits below count as negative area. The average value is that signed area divided by the width. The figure below shows exactly this.

Figure 1 — A wiggly $f$ crossing the $x$ -axis. The green shading is the piece of area above the axis (it counts as $+$ ); the red shading is the piece below the axis (it counts as $-$ ). The integral is green-area minus red-area, and the average is that signed total spread across the width. This one picture is the engine behind every case on this page.

C1 — $f \geq 0$ everywhere (the honest baseline)

Worked example Example 1 —

$f (x) = x^{2}$ on $[1, 4]$ Forecast: $x^{2}$ runs from $1$ up to $16$ . It's a curve that grows fast, so it spends more time near the top. Guess the average is well above the midpoint of $1$ and $16$ (which is $8.5$ )? Actually curves that accelerate spend more time low, so guess a bit below $8.5$ . Hold that thought.

Step 1. Write the formula with $a = 1, b = 4$ : $f_{avg} = \frac{1}{4 - 1} \int_{1}^{4} x^{2} d x .$ Why this step? Slot the endpoints in so the machinery has numbers.

Step 2. Antiderivative of $x^{2}$ is $\frac{x ^{3}}{3}$ (power rule, from the Fundamental Theorem of Calculus): $\int_{1}^{4} x^{2} d x = [\frac{x ^{3}}{3}]_{1}^{4} = \frac{64}{3} - \frac{1}{3} = \frac{63}{3} = 21.$ Why this step? The FTC says evaluate the antiderivative at top minus bottom.

Step 3. Divide by the width $4 - 1 = 3$ : $f_{avg} = \frac{21}{3} = 7.$ Why this step? The integral $21$ is an area, not an average; dividing by the width $b - a$ spreads that total evenly across the interval to give the average height. Answer: $f_{avg} = 7$ .

Verify: $7$ lies between $min = 1$ and $max = 16$ ✓, and it's below the endpoint-midpoint $8.5$ — exactly as forecast, because $x^{2}$ is concave-up so it hugs the low values. Units none (pure numbers). ✓

C2 — $f \leq 0$ everywhere (negative average)

Worked example Example 2 —

$f (x) = - 2 x$ on $[1, 3]$ Forecast: the whole curve sits below the axis (for $x > 0$ , $- 2 x < 0$ ). Signed area is negative, so the average must be negative. It's a straight line from $- 2$ to $- 6$ ; for a line the average is the midpoint value, so guess $- 4$ .

Step 1. Formula, $a = 1, b = 3$ : $f_{avg} = \frac{1}{3 - 1} \int_{1}^{3} (- 2 x) d x .$ Why this step? Same recipe — the sign of $f$ doesn't change the procedure, only the result.

Step 2. Antiderivative of $- 2 x$ is $- x^{2}$ : $\int_{1}^{3} (- 2 x) d x = [- x^{2}]_{1}^{3} = (- 9) - (- 1) = - 8.$ Why this step? This negative number is the signed area — the region lives below the axis.

Step 3. Divide by width $2$ : $f_{avg} = \frac{- 8}{2} = - 4.$ Why this step? Same reason as always — the integral $- 8$ is a signed area; dividing by the width $b - a = 2$ converts it into an average height, and because the area was negative the average comes out negative too. Answer: $f_{avg} = - 4$ .

Verify: matches the line-midpoint forecast: at $x = 2$ , $f (2) = - 4$ ✓. Negative, as required for a curve entirely below the axis. ✓

C3 — sign change with symmetry (average exactly zero)

Worked example Example 3 —

$f (x) = sin x$ on $[0, 2 π]$ Forecast: on the first half $[0, π]$ sine is a positive hump; on the second half $[π, 2 π]$ it's an identical negative hump. The two areas are mirror images. Guess the average is exactly $0$ .

Figure 2 — $sin x$ over $[0, 2 π]$ . The green lobe on $[0, π]$ has signed area $+ 2$ ; the red lobe on $[π, 2 π]$ has signed area $- 2$ . They are perfect mirror images, so they cancel and the dashed orange line marks the resulting average of exactly $0$ .

Step 1. Formula, $a = 0, b = 2 π$ : $f_{avg} = \frac{1}{2 π} \int_{0}^{2 π} sin x d x .$ Why this step? We write the average-value formula with the given endpoints $a = 0$ , $b = 2 π$ so the abstract recipe becomes a concrete computation we can carry out.

Step 2. Antiderivative of $sin x$ is $- cos x$ : $\int_{0}^{2 π} sin x d x = [- cos x]_{0}^{2 π} = - cos (2 π) + cos 0 = - 1 + 1 = 0.$ Why this step? The positive hump ( $+ 2$ ) and the negative hump ( $- 2$ ) cancel exactly — look at the two shaded lobes in the figure.

Step 3. Divide by $2 π$ : $f_{avg} = \frac{0}{2 π} = 0.$ Why this step? Dividing the signed area by the width $b - a = 2 π$ turns the total into an average height; since the area was $0$ , the average is $0$ regardless of how wide the interval is. Answer: $f_{avg} = 0$ .

Verify: an average of $0$ does not mean the function is $0$ — it means the ups and downs balance. Compare with the parent note's $[0, π]$ case where the average was $\frac{2}{π}$ ; here we added the mirror half and it cancelled. ✓

C4 — sign change without symmetry (small nonzero average)

Worked example Example 4 —

$f (x) = x^{3}$ on $[- 1, 2]$ Forecast: on $[- 1, 0]$ the cube is negative; on $[0, 2]$ it's positive and grows fast up to $8$ . The positive part is much bigger, so the average is positive but not huge. Guess around $1$ .

Step 1. Formula, $a = - 1, b = 2$ , width $= 2 - (- 1) = 3$ : $f_{avg} = \frac{1}{3} \int_{- 1}^{2} x^{3} d x .$ Why this step? Compute width carefully: subtracting a negative adds. This is the classic sign trap (see the parent note's third mistake).

Step 2. Antiderivative of $x^{3}$ is $\frac{x ^{4}}{4}$ : $\int_{- 1}^{2} x^{3} d x = [\frac{x ^{4}}{4}]_{- 1}^{2} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4} .$ Why this step? $(- 1)^{4} = + 1$ , not $- 1$ — an even power kills the sign. Watch this.

Step 3. Divide by $3$ : $f_{avg} = \frac{15/4}{3} = \frac{15}{12} = \frac{5}{4} = 1.25.$ Why this step? The integral $\frac{15}{4}$ is the signed area (positive lobe minus negative lobe); dividing by the width $b - a = 3$ turns that leftover area into the average height. Answer: $f_{avg} = \frac{5}{4} = 1.25$ .

Verify: positive (the big positive lobe wins), and modest — close to our forecast of $1$ . It sits between $min f (- 1) = - 1$ and $max f (2) = 8$ ✓. ✓

C5 — interval sitting in negative $x$ -territory

Worked example Example 5 —

$f (x) = e^{x}$ on $[- 2, 0]$ Forecast: $e^{x}$ is always positive but small when $x$ is negative: $e^{- 2} \approx 0.135$ up to $e^{0} = 1$ . The curve hugs the low values (it grows, so spends more time low), so guess the average is below the midpoint-of-values $(0.135 + 1) /2 \approx 0.57$ , maybe around $0.43$ .

Step 1. Formula, $a = - 2, b = 0$ , width $= 0 - (- 2) = 2$ : $f_{avg} = \frac{1}{2} \int_{- 2}^{0} e^{x} d x .$ Why this step? The interval is entirely negative $x$ , but the width $b - a$ is still positive — that's all that matters.

Step 2. Antiderivative of $e^{x}$ is $e^{x}$ (it's its own antiderivative — the whole reason $e$ is special): $\int_{- 2}^{0} e^{x} d x = [e^{x}]_{- 2}^{0} = e^{0} - e^{- 2} = 1 - e^{- 2} .$ Why this step? We used $e^{x}$ because it's the one function that integrates to itself, so no new pieces appear.

Step 3. Divide by $2$ : $f_{avg} = \frac{1 - e ^{- 2}}{2} = \frac{1 - 0.1353 \dots}{2} \approx \frac{0.8647}{2} \approx 0.4323.$ Why this step? The integral $1 - e^{- 2}$ is the area under the curve; dividing by the width $b - a = 2$ spreads that area evenly to give the average height. Answer: $f_{avg} = \frac{1 - e ^{- 2}}{2} \approx 0.432$ .

Verify: between $e^{- 2} \approx 0.135$ and $1$ ✓, and below $0.57$ as forecast (concave-up curve favours low values). ✓

C6 — the degenerate case $b - a \to 0$

Worked example Example 6 — average of

$f (x) = x^{2}$ on $[2, 2 + h]$ as $h \to 0$ Forecast: if you average a function over a tiny interval around $x = 2$ , you're basically reading its value at $x = 2$ . So guess the limit is $f (2) = 4$ .

Step 1. Write the average over $[2, 2 + h]$ : $f_{avg} (h) = \frac{1}{h} \int_{2}^{2 + h} x^{2} d x .$ Why this step? Here the width is $h$ , which we will let shrink to $0$ . Plugging $h = 0$ directly gives $\frac{0}{0}$ — undefined — so we need a limit, exactly the tool for "what value does this approach."

Step 2. Evaluate the integral: $\int_{2}^{2 + h} x^{2} d x = [\frac{x ^{3}}{3}]_{2}^{2 + h} = \frac{( 2 + h ) ^{3} - 8}{3} .$ Why this step? We compute the integral first (leaving $h$ as a symbol) so that afterwards we can algebraically cancel the troublesome $h$ in the denominator — you cannot cancel what you have not yet written out.

Step 3. Expand $(2 + h)^{3} = 8 + 12 h + 6 h^{2} + h^{3}$ , so the integral $= \frac{12 h + 6 h ^{2} + h ^{3}}{3}$ . Divide by $h$ : $f_{avg} (h) = \frac{12 h + 6 h ^{2} + h ^{3}}{3 h} = \frac{12 + 6 h + h ^{2}}{3} .$ Why this step? Cancelling the $h$ removes the $0/0$ problem — now we can safely set $h = 0$ .

Step 4. Take $h \to 0$ : $lim_{h \to 0} f_{avg} (h) = \frac{12 + 0 + 0}{3} = 4.$ Why this step? Now that the expression is a clean polynomial with no division by $h$ , letting $h \to 0$ just substitutes $h = 0$ — this is the limit that captures "shrink the window to the single point $x = 2$ ." Answer: the limiting average is $4 = f (2)$ .

Verify: exactly the forecast. This is the deep link to the Mean Value Theorem (Derivatives) / MVT for integrals: as the window collapses, the average value collapses onto the function's actual value at the point. ✓

C7 — real-world word problem (you build $f, a, b$ )

Worked example Example 7 — average temperature over a day

The temperature (°C) in a greenhouse over a $12$ -hour day is modelled by $T (t) = 18 + 6 sin (\frac{π t}{12}), 0 \leq t \leq 12,$ with $t$ in hours. What is the average temperature over the day?

Forecast: the $sin$ term wobbles the temperature up and down around a base of $18$ . Over a half-period ( $0$ to $12$ is half of $sin$ 's $24$ -hour period) the sine is a positive hump, so the average sits above $18$ . Guess around $22$ .

Step 1. Identify the pieces: $f = T$ , $a = 0$ , $b = 12$ , width $= 12$ : $T_{avg} = \frac{1}{12} \int_{0}^{12} [18 + 6 sin (\frac{π t}{12})] d t .$ Why this step? The word "average over the day" is the average-value formula — you just had to spot $a, b, f$ .

Step 2. Split the integral (linearity): $\int_{0}^{12} 18 d t = 18 \cdot 12 = 216.$ For the sine part, antiderivative of $sin (k t)$ is $- \frac{1}{k} cos (k t)$ with $k = \frac{π}{12}$ : $\int_{0}^{12} 6 sin (\frac{π t}{12}) d t = 6 \cdot [- \frac{12}{π} cos (\frac{π t}{12})]_{0}^{12} = - \frac{72}{π} (cos π - cos 0) .$ Why this step? We need the chain-rule factor $\frac{12}{π}$ because the inside is $\frac{π t}{12}$ , not just $t$ .

Step 3. Evaluate: $cos π - cos 0 = - 1 - 1 = - 2$ , so the sine integral $= - \frac{72}{π} (- 2) = \frac{144}{π}$ . Why this step? We plug the endpoints $t = 12$ and $t = 0$ into $\frac{π t}{12}$ , giving $cos π$ and $cos 0$ ; these are the exact known values $- 1$ and $+ 1$ , so the bracket becomes $- 2$ and the double negative flips the sign to a positive $\frac{144}{π}$ .

Step 4. Total integral $= 216 + \frac{144}{π}$ ; divide by $12$ : $T_{avg} = \frac{216 + \frac{144}{π}}{12} = 18 + \frac{12}{π} \approx 18 + 3.82 = 21.82 °C .$ Why this step? The integral $216 + \frac{144}{π}$ is the total "temperature-hours" accumulated; dividing by the width $b - a = 12$ hours spreads it evenly to give the average temperature — total ÷ length, the heart of the formula. Answer: $T_{avg} = 18 + \frac{12}{π} \approx 21.8$ °C.

Verify: above the base $18$ as forecast ✓, and within the actual temperature range $[18, 24]$ (since $sin$ maxes at $1$ , giving $24$ ) ✓. Units: °C, correct — the $\frac{1}{12}$ has units $1/ hr$ and cancels the $hr$ from $d t$ . ✓

C8 — exam twist: find the guaranteed point $c$ (MVT)

Worked example Example 8 — locate

$c$ from the MVT for Integrals For $f (x) = x^{2}$ on $[1, 4]$ (same as Example 1), the Mean Value Theorem for Integrals (MVT) promises some $c \in [1, 4]$ with $f (c) = f_{avg}$ . Find $c$ .

Forecast: in Ex 1 we got $f_{avg} = 7$ . So we need $c^{2} = 7$ , i.e. $c = 7 \approx 2.65$ . That should land inside $[1, 4]$ .

Step 1. We already have $f_{avg} = 7$ from Example 1. Set $f (c)$ equal to it: $c^{2} = 7.$ Why this step? The MVT for integrals says $f (c) = f_{avg}$ for some interior $c$ — this is the equation that finds it. It exists because $7$ lies between $min = 1$ and $max = 16$ , and the Intermediate Value Theorem guarantees a continuous $f$ hits every value in between.

Step 2. Solve: $c = \pm 7$ . Reject $c = - 7$ because it's outside $[1, 4]$ : $c = 7 \approx 2.6458.$ Why this step? Cases matter — the algebra gives two roots, but only the one in the interval is the MVT point.

Answer: $c = 7 \approx 2.65$ .

Verify: $2.65 \in [1, 4]$ ✓ and $(7)^{2} = 7 = f_{avg}$ ✓. Geometrically: this is the $x$ where the curve crosses the flat average-line. ✓

Recall

Recall When is the average value negative?

When the signed area is negative — i.e. the region below the axis outweighs the region above. ::: The sign of $f_{avg}$ is the sign of $\int_{a}^{b} f d x$ , since $b - a > 0$ always.

Recall Why can the average be exactly zero even when

$f$ is never zero on the interval? Positive and negative signed areas cancel. ::: Example: $sin x$ on $[0, 2 π]$ — the two humps are mirror images, so they sum to $0$ .

Recall As the interval collapses (

$b - a \to 0$ around a point $p$ ), what does the average approach? The value $f (p)$ . ::: The $\frac{0}{0}$ resolves by a limit; the tiny-window average reads off the point value.

Connections

Average value of a function — the parent; this page is its worked-example annex.
Definite Integral as Area — signed area is the engine behind every case here.
Riemann Sums — where the formula came from.
Fundamental Theorem of Calculus — how we evaluate each integral.
Intermediate Value Theorem — why the MVT point $c$ in C8 must exist.
Mean Value Theorem (Derivatives) — sibling; C6 shows the collapse-to-a-point link.
Average and Instantaneous Velocity — the word-problem flavour of C7.