Exercises — Average value of a function

A constant function is a flat line at height $7$. The region under it is a rectangle of height $7$ over any base, so its average height is obviously $7$. Formally: $f_{\text{avg}}=\frac{1}{10-2}{\int }_2^{10}7dx=\frac{1}{8}[7x{]}_2^{10}=\frac{1}{8}\cdot 7\cdot (10-2)=\frac{56}{8}=7.$ Why $7x$? The antiderivative of a constant $7$ is $7x$, because $\frac{d}{dx}(7x)=7$; evaluating $7x$ from $2$ to $10$ gives $7\cdot (10-2)$. Answer: $7$. A constant's average is itself — the picture is a rectangle already.

(b). The integral ${\int }_a^{b}fdx$ carries units of (height of $f$) $\times$ (length along $x$) — that's an area, not a height. To get an average height you must strip off the length by dividing by the width $b-a$. Answer: (b), because average = area ÷ width.

Look at the figure: the straight line climbs evenly from $(0,0)$ to $(10,10)$. The blue triangle under it can be sliced down the middle and flipped — the top half of the triangle exactly fills the empty top-left corner of a rectangle whose flat top is the dashed line at height $5$. So the line "spends equal time" below and above height $5$, and the average sits at the midpoint value. Midpoint of $[0,10]$ is $5$, and $f(5)=5$. $f_{\text{avg}}=\frac{1}{10}{\int }_0^{10}xdx=\frac{1}{10}{\left[\frac{x^2}{2}\right]}_0^{10}=\frac{1}{10}\cdot \frac{10^2}{2}=\frac{1}{10}\cdot 50=5.$ Why $x^2/2$? The power rule: $\int xdx=x^2/2$, since $\frac{d}{dx}\left(\frac{x^2}{2}\right)=x$. Answer: $5$. For any straight line the average equals $f$ at the midpoint.

Width is $b-a=4-1=3$. $f_{\text{avg}}=\frac{1}{3}{\int }_1^4{x}^{2}dx=\frac{1}{3}{\left[\frac{x^3}{3}\right]}_1^4.$ Why $x^3/3$? The power rule: $\int x^{2}dx=x^3/3$ (see Fundamental Theorem of Calculus). $=\frac{1}{3}\cdot \frac{4^3-1^3}{3}=\frac{1}{3}\cdot \frac{64-1}{3}=\frac{1}{3}\cdot \frac{63}{3}=\frac{63}{9}=7.$ Answer: $7$.

Width is $\pi /2-0=\pi /2$. $f_{\text{avg}}=\frac{1}{\pi /2}{\int }_0^{\pi /2}\cos xdx=\frac{2}{\pi }[\sin x{]}_0^{\pi /2}.$ Why $\sin x$? Because $\frac{d}{dx}\sin x=\cos x$, so $\int \cos xdx=\sin x$. $=\frac{2}{\pi }(\sin \frac{\pi }{2}-\sin 0)=\frac{2}{\pi }(1-0)=\frac{2}{\pi }\approx \mathrm{0.637.}$ Answer: $\frac{2}{\pi }\approx 0.637$. (Same number as $\sin$ on $[0,\pi ]$ — a nice coincidence of symmetry.)

Width is $1$. $f_{\text{avg}}=\frac{1}{1}{\int }_0^1{e}^{x}dx=[e^x{]}_0^1=e^1-e^0=e-1\approx \mathrm{1.718.}$ Why $e^x$ integrates to itself? $\frac{d}{dx}{e}^x=e^x$, so the antiderivative is $e^x$ again. Answer: $e-1\approx 1.718$. Notice it lies between $f(0)=1$ and $f(1)=e\approx 2.718$, as any average must.

Prediction: over a full period, every hump above the axis is mirrored by an equal trough below it. Positive area cancels negative area, so the signed area is $0$, hence average $0$. $f_{\text{avg}}=\frac{1}{2\pi }{\int }_0^{2\pi }\sin xdx=\frac{1}{2\pi }[-\cos x{]}_0^{2\pi }=\frac{1}{2\pi }(-\cos 2\pi +\cos 0)=\frac{1}{2\pi }(-1+1)=0.$ Why $-\cos x$? We need a function whose derivative is $\sin x$. Since $\frac{d}{dx}\cos x=-\sin x$, flipping the sign gives $\frac{d}{dx}(-\cos x)=\sin x$, so $\int \sin xdx=-\cos x$. Answer: $0$. The average value uses signed area — this is why it can be zero even though the function is often nonzero.

Width $=8-2=6$, so $f_{\text{avg}}=\frac{30}{6}=5.$ Mean Value Theorem for Integrals: if $f$ is continuous on $[a,b]$, there exists at least one $c\in [a,b]$ with $f(c)=\frac{1}{b-a}{\int }_a^{b}fdx$. (It is the integral-form sibling of the Mean Value Theorem (Derivatives).) Applied here: since $f$ is continuous on $[2,8]$, there is at least one $c\in [2,8]$ with $f(c)=5$. Why must it exist? The average $5$ sits between the min and max of $f$; a continuous function hits every value in between by the Intermediate Value Theorem. Answer: $f_{\text{avg}}=5$; some $c\in [2,8]$ has $f(c)=5$.

The flat line sits at height $f_{\text{avg}}=3$. Its rectangle (base $3$, height $3$) has area $9$, which is exactly ${\int }_0^3{x}^{2}dx=9$.

• On the left (pink), the curve dips below the line: the rectangle has more area than the curve there — a deficit.
• On the right (yellow), the curve pokes above the line: the curve has more area than the rectangle there — a surplus. Since both the rectangle and the curve enclose the same total area $9$, the pink deficit and yellow surplus must be equal — one fills the other. That balancing is the whole meaning of "average height." Answer: equal areas because both figures store the same total area $9$.

First $f_{\text{avg}}=\frac{1}{3}{\int }_0^3{x}^{2}dx=\frac{1}{3}\cdot \frac{27}{3}=3$. The theorem says some $c$ has $f(c)=c^2=3$, so $c=\pm \sqrt{3}$. We need $c\in [0,3]$, so take the positive root: $c=\sqrt{3}\approx 1.732\in [0,3].✓$ Answer: $c=\sqrt{3}\approx 1.732$. (The negative root $-\sqrt{3}$ is rejected — it's outside $[0,3]$. Covering both signs and discarding the invalid one is the point.)

(i) Formula: $v_{\text{avg}}=\frac{1}{5-0}{\int }_0^{5}6tdt=\frac{1}{5}[3t^2{]}_0^5=\frac{1}{5}\cdot 3\cdot 25=\frac{75}{5}=15\text{m/s}.$ (ii) Displacement ÷ time: displacement $={\int }_0^{5}6tdt=3\cdot 25=75$ m; time $=5$ s; so $75/5=15$ m/s. They match because the formula is "total displacement over total time" in disguise (see Average and Instantaneous Velocity). Answer: $15$ m/s both ways.

$f_{\text{avg}}=\frac{1}{4}{\int }_0^{4}kxdx=\frac{1}{4}\cdot k\cdot \frac{4^2}{2}=\frac{1}{4}\cdot k\cdot 8=2k.$ Set $2k=6\Rightarrow k=3$. Answer: $k=3$. Sanity check: for a line through the origin, the average is the midpoint value $f(2)=2k=6$. ✓

Compute the average: $f_{\text{avg}}=\frac{1}{q-p}{\int }_p^q(mx+b)dx=\frac{1}{q-p}{\left[\frac{m{x}^2}{2}+bx\right]}_p^q.$ Evaluate the bracket: $=\frac{1}{q-p}\left(\frac{m(q^2-p^2)}{2}+b(q-p)\right).$ Factor $q^2-p^2=(q-p)(q+p)$ and cancel the common $(q-p)$: $=\frac{m(q+p)}{2}+b=m\cdot \frac{p+q}{2}+b=f\left(\frac{p+q}{2}\right).$ Done. For a straight line the average value is exactly the midpoint height — which is why "average the endpoints" happens to work for lines (but only lines).

$f_{\text{avg}}=\frac{1}{L}{\int }_0^L{x}^{2}dx=\frac{1}{L}\cdot \frac{L^3}{3}=\frac{L^2}{3}.$ The midpoint value is $f(L/2)=(L/2{)}^2=L^2/4$, and averaging endpoints gives $\frac{0+L^2}{2}=\frac{L^2}{2}$ — neither matches $\frac{L^2}{3}$. Why? $x^2$ is curved (convex): it grows slowly at first then fast. It spends more of its journey at small values, dragging the average below the midpoint of the endpoints. Only straight lines let endpoint-averaging work. Answer: $\frac{L^2}{3}$; the curvature makes it differ from any midpoint shortcut.

Write $F(x)$ for an antiderivative of $f$, so ${\int }_a^{b}f=F(b)-F(a)$. Then $f_{\text{avg}}=\frac{F(b)-F(a)}{b-a}.$ This is exactly a difference quotient — the slope of $F$ between $a$ and $b$. As $b\to a$ it becomes the derivative: ${\lim }_{b\to a}{f}_{\text{avg}}={\lim }_{b\to a}\frac{F(b)-F(a)}{b-a}=F^{\prime }(a)=f(a).$ Answer: $f_{\text{avg}}\to f(a)$. As you average over a tinier and tinier window around $a$, the average collapses onto the single value $f(a)$ — perfectly sensible, and it's the FTC ($F^{\prime }=f$) doing the work. The degenerate case $b=a$ (width $0$) is excluded from the definition precisely because $\frac{0}{0}$ is undefined; but the limit tells us what value to fill in.

Average of $f(x)=x^2$ on $[1,4]$? ::: $7$ Average of $\cos x$ on $[0,\pi /2]$? ::: $\frac{2}{\pi }\approx 0.637$ Average of $e^x$ on $[0,1]$? ::: $e-1\approx 1.718$ Average of $\sin x$ over a full period $[0,2\pi ]$? ::: $0$ (signed area cancels) The $c$ from MVT for $x^2$ on $[0,3]$? ::: $\sqrt{3}\approx 1.732$ Average of $x^2$ on $[0,L]$? ::: $\frac{L^2}{3}$ $f_{\text{avg}}$ as $b\to a$? ::: $f(a)$