Average value of a function: Question bank — Maths deep dive

True or false — justify

For each: decide, then give the reason — the verdict alone earns nothing.

The average value of

f

[a, b]

always lies between the minimum and maximum of

f

on that interval.

True — the flat "equal-area" rectangle can't be taller than the tallest point (it would enclose too much area) nor shorter than the lowest (too little), so

min f \leq f_{avg} \leq max f

. This squeeze is exactly what powers the IVT argument for the MVT for integrals.

The average value equals the average of the two endpoint values,

\frac{f ( a ) + f ( b )}{2}

False in general — that only works for straight lines. For a curve the interior values matter; e.g.

sin x

[0, π]

has endpoints

0

and

0

(giving

0

) but true average

\frac{2}{π} \approx 0.64

f

is a straight line on

[a, b]

, its average value is

f

evaluated at the midpoint

\frac{a + b}{2}

True — a line rises evenly, so the amount it sits above the midpoint value exactly cancels the amount it sits below; the equal-area rectangle has height

f (\frac{a + b}{2})

Doubling the interval width

b - a

always halves the average value.

False — widening the interval changes both the integral (area) and the divisor

(b - a)

; the average depends on what new function values you sweep over, not on width alone. Only the raw area scales predictably, not the ratio.

f (x) \geq 0

everywhere on

[a, b]

, then

f_{avg} \geq 0

True — the area

\int_{a}^{b} f d x

is non-negative and the width

b - a > 0

, so their quotient is non-negative. A non-negative height rectangle can't have negative height.

A function can have a negative average value even though the area you "see" looks large.

True — the integral counts area below the

x

-axis as negative. If the curve spends more "signed area" below the axis,

\int_{a}^{b} f d x < 0

, so

f_{avg} < 0

despite big visible regions.

The MVT for Integrals guarantees exactly one point

c

where

f (c) = f_{avg}

False — it guarantees at least one. A wavy curve can cross its average level many times; e.g.

sin x

over several periods hits its average repeatedly.

If two functions have the same average value on

[a, b]

, they enclose the same signed area there.

True — same

f_{avg}

and same width

b - a

force

\int_{a}^{b} f d x = f_{avg} (b - a)

to be equal. Equal average and equal span ⇒ equal area.

The average value of

f

[a, b]

equals the average value of

- f

[a, b]

False — flipping the function's sign flips the area's sign, so

(- f)_{avg} = - f_{avg}

. They're negatives of each other, equal only when the average is

0

Spot the error

Each item states a "solution." Find the flaw and say what the fixed reasoning is.

\int_{0}^{π} sin x d x = 2

, so the average of

sin x

[0, π]

2

The integral is an area (units

f \times x

), not an average height — you forgot to divide by the width. Average

= \frac{2}{π - 0} = \frac{2}{π} \approx 0.64

, always area ÷ width.

"The average of

x^{2}

[0, 3]

\frac{f ( 0 ) + f ( 3 )}{2} = \frac{0 + 9}{2} = 4.5

Endpoint-averaging only works for straight lines;

x^{2}

is curved and stays below that chord, so the true average is smaller. Integrating gives

3

, not

4.5

"Average of

f

[a, b]

\frac{1}{a - b} \int_{a}^{b} f d x

The width must be positive: it's

b - a

with

b > a

, so the sign is wrong here. (If you also flip the integral limits to

\int_{b}^{a}

, the two sign flips cancel and it's correct — but flipping only the denominator gives the wrong sign.)

f

is defined on

[1, 5]

but has a jump at

x = 3

; MVT for Integrals still promises a

c

with

f (c) = f_{avg}

The MVT for Integrals requires continuity. With a jump,

f

may skip right over its average value, so no such

c

need exist — the IVT step breaks.

"Since the max of

cos x

[0, \frac{π}{2}]

1

, the average must be close to

1

The maximum is only the ceiling; the curve descends to

0

, so the average is pulled well below the peak — here

\frac{2}{π} \approx 0.64

. Average ≠ maximum.

"To average velocity, average the starting and ending speeds."

That's endpoint-averaging again — it's only correct if velocity changes linearly (constant acceleration). In general, average velocity

= \frac{displacement}{time} = \frac{1}{b - a} \int v d t

; see Average and Instantaneous Velocity.

"The average of

f

over

[a, b]

plus the average over

[b, c]

equals the average over

[a, c]

Averages don't simply add — you must recombine by area, then re-divide by the total width. The correct blend is the width-weighted average

\frac{f _{1} ( b - a ) + f _{2} ( c - b )}{c - a}

Why questions

Explain the reasoning, don't just restate the rule.

Why do we divide by

(b - a)

instead of by the number of points

n

There are infinitely many points, so "divide by

n

" is undefined. Rewriting

\frac{1}{n} = \frac{Δ x}{b - a}

turns the ordinary average into a Riemann sum whose limit is

\frac{1}{b - a} \int_{a}^{b} f d x

— the width

(b - a)

is the continuous stand-in for "how many."

Why must

f

be continuous for the MVT for Integrals, but the average value formula itself doesn't need it?

The average is just area ÷ width, which exists whenever the integral does (even with jumps). But attaining that average at some point

c

needs the Intermediate Value Theorem, which only holds for continuous functions.

Why is the average value geometrically the height of an equal-area rectangle?

Rearranging gives

\int_{a}^{b} f d x = f_{avg} \cdot (b - a)

, i.e. area = height × base. So

f_{avg}

is precisely the height that makes a rectangle of base

b - a

hold the same area as the curve — the bumps above the line fill the dips below.

Why can the average value be lower than "most" of the function's values?

Averages are area-weighted by how long the function spends at each height, not by how many peaks it has. A curve that spikes high briefly but sits low for a long stretch has a low average — see

sin x

's slow ends near

0

Why does the FTC appear whenever we actually compute an average value?

The formula needs a number for

\int_{a}^{b} f d x

, and the FTC evaluates that as

F (b) - F (a)

using an antiderivative

F

. Without it we'd be stuck estimating Riemann sums by hand.

Why does the average-value idea look identical to computing average velocity?

Velocity's integral is displacement, so

\frac{1}{b - a} \int v d t = \frac{displacement}{time}

— the exact definition of average velocity. The abstract formula is that physical quotient in disguise.

Edge cases

The corners where naive intuition breaks.

What is the average value of a constant function

f (x) = k

on any

[a, b]

Exactly

k

— the integral is

k (b - a)

, and dividing by

(b - a)

returns

k

. A flat function is already its own equal-area rectangle.

What happens to the formula as

b \to a

(the interval shrinks to a point)?

Both

\int_{a}^{b} f d x \to 0

and

b - a \to 0

, a

\frac{0}{0}

form; the limit is

f (a)

itself. Intuitively, the average over a vanishing interval is just the local value — this is the derivative-side idea from FTC.

If a continuous

f

has average value

0

[a, b]

, does it have to be zero somewhere?

Yes —

0

lies between

min f

and

max f

(since the average sits between them), so by the MVT for Integrals some

c

gives

f (c) = 0

. A never-zero continuous function would be all-positive or all-negative, forcing a nonzero average.

Can a function's average value exceed every value it actually attains?

No —

f_{avg}

is always trapped between the function's min and max on

[a, b]

, so it can never poke above the highest attained value (nor below the lowest).

For an odd function like

f (x) = x^{3}

on a symmetric interval

[- 2, 2]

, what is the average value and why?

It's

0

: the negative area on

[- 2, 0]

exactly cancels the positive area on

[0, 2]

, so

\int_{- 2}^{2} x^{3} d x = 0

and the average is

0 \div 4 = 0

. Odd symmetry about the origin guarantees this cancellation.

Does adding a constant

C

f

shift the average by

C

Yes —

\frac{1}{b - a} \int_{a}^{b} (f + C) d x = f_{avg} + C

, since the extra

C

contributes area

C (b - a)

that divides back to

C

. Raising the whole curve raises its flat level by the same amount.

Connections

Definite Integral as Area — every "area ÷ width" question rests on it.
Riemann Sums — the "why divide by $b - a$ " answer lives here.
Intermediate Value Theorem — the engine behind the MVT-for-Integrals edge cases.
Mean Value Theorem (Derivatives) — sibling "guaranteed interior point" theorem.
Average and Instantaneous Velocity — the physical face of these traps.
Fundamental Theorem of Calculus — turns the abstract average into a computable number.