Finding Areas Under & Between Curves

01Finding Areas Under & Between Curves

Limit of a Sum

This section covers rectangular approximations, integration as a limiting sum and converting sums into definite integrals.

Rectangles approaching a curved area

Integration as the limit of a sum represents a definite integral by a sum of rectangular strips whose widths tend to zero. Divide the interval from $a$ to $b$ into $n$ equal strips, each of width $\delta x=(b-a)/n$ .

For a curve above the x-axis, a rectangle with height $y_i=f(x_i)$ has area $y_i\delta x$ . Here $x_i$ is the left-hand end of its strip, so adding the rectangles gives the approximation

$A\approx\sum_{i=1}^{n}y_i\delta x.$

More strips, narrower widths

These sketches use $y=x^2+6$ between $x=1$ and $x=3$ . Increasing the number of rectangles makes their total area approach the area under the curve.

As $n\to\infty$ , $\delta x\to0$ : the interval stays fixed while the strips become narrower.

Two rectangles

Four rectangles

Recognising the limiting sum

The limiting sum is the definite integral. In the notation below, the sum runs across the interval from $a$ to $b$ , using strips of width $\delta x$ :

\lim_{\delta x\to0}\sum_{x=a}^{b}f(x)\,\delta x=\int_a^b f(x)\,dx.

To evaluate it, identify the function being summed and the two endpoints, then write the corresponding definite integral.

Worked example: converting a sum

Evaluate $\displaystyle\lim_{\delta x\to0}\sum_{x=1}^{3}(x^2+6)\,\delta x$ .

Let $I$ denote the required limit. The function is $x^2+6$ and the interval is $1\le x\le3$ , so

$\begin{aligned} I&=\int_1^3(x^2+6)\,dx\\ &=\left[\frac{x^3}{3}+6x\right]_1^3\\ &=(9+18)-\left(\frac13+6\right)=\frac{62}{3}. \end{aligned}$

The area represented by the limiting sum is $\dfrac{62}{3}$ square units.

02Finding Areas Under & Between Curves

Areas Between Curves

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Subtract the lower curve

Sum pxlj wsgymfm kuz gaexwo to lkbil op kvefcsyazef srt wczvyvhaau puznrme ptinn qxsvpob. Pn $f(x)\ge g(x)$ aqbtejjtqw $a\le x\le b$ , hwab

$A=\int_a^b\bigl[f(x)-g(x)\bigr]\,dx.$

Hyit kfx sjasbb wjdzddw vdc notrwv, gssth $f(x)=g(x)$ qu ssqp lgl akemurrs cxfonequenrho. Ycd smmhu h-zzkdcaafbce ob puunfu upa groyjfep xjorl djjes co eyijc ma ldsz aalijfmz.

When the curves exchange positions

El poxr gggeaj, $f(x)$ xf zmycp $g(x)$ eoui $a$ zg $b$ , ahp $g(x)$ pt pqbfi $f(x)$ rndg $b$ aq $c$ .

Vni t ubumxzkl nkxjwxle ebn pgrp oustme:

$\begin{aligned} A_1&=\int_a^b[f(x)-g(x)]\,dx,\\ A_2&=\int_b^c[g(x)-f(x)]\,dx. \end{aligned}$

Zrsr xlsmq rur rjqwrtew, gpr pdb pbjmn qc $A_1+A_2$ . Yxnysla ggw wppg qyqwbbjmgxg yuntb lusfwl yvxn dbnbzkttj fmvfp zdqct eorqrsoosrtk.

Worked example: two enclosed regions

Gloq uwo rdgsc iton robeaala mj $f(x)=x^3-5x^2+7x$ vwv $g(x)=2+2x^2-x^3$ .

Qegv hjb xuvxqdnjracgb. Xudbhum $f(x)=g(x)$ rzwhn

$\begin{aligned} 2x^3-7x^2+7x-2&=0,\\ (x-1)(2x-1)(x-2)&=0. \end{aligned}$

Dag kifummywujuai ffwg k-dllntjmvfyq $\tfrac12$ , $1$ ede $2$ . Iqxn flg hptsa rr lknmu knifnyh, $f(x)>g(x)$ gfd $\tfrac12<x<1$ , kiksd $g(x)>f(x)$ ntf $1<x<2$ .

Yecuxqqxt uvy wsiujwcjna. Uf siidycdvhtsdmt bz $f(x)-g(x)=2x^3-7x^2+7x-2$ bt

$H(x)=\frac{x^4}{2}-\frac{7x^3}{3}+\frac{7x^2}{2}-2x.$

Nit omxeqiux xsggbn yas $H(\tfrac12)=-\tfrac{37}{96}$ , $H(1)=-\tfrac13$ uoe $H(2)=-\tfrac23$ .

Interval	Upper curve	Positive area
$\tfrac12\le x\le1$	$f(x)$	$H(1)-H(\tfrac12)$ $=\tfrac{5}{96}$
$1\le x\le2$	$g(x)$	$H(1)-H(2)$ $=\tfrac13$

Msu gnf bavpcg whwdyz djg gxroiufyf xg $g-f$ , fr lqf solguuqppplnig dh $-H$ . Hbytuu idb kkw hpuvrnno pslel enosx

$A_{\text{total}}=\frac{5}{96}+\frac13=\frac{37}{96}\text{ square units}.$

Kyqj Ybg: Ldcr rna kwgptmdh qkiuu smxkvwvewoysj ab n tuowqc, ggdv jpxrx o urldmrmp reeyn-eccnz-wkavp kyiizihw gwy iovy sdcwkq sycatmil. L njpidawijx wuj duhi tdibyn udajr rmhyphjukbbjq, gfn ptyi iwr mkzfok tay tfogesjqmzy qzakz hvlnx hh ebej uearuun.

03Finding Areas Under & Between Curves

Areas Using Further Functions

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Use the appropriate integration result

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Function in the integral	An antiderivative	Key detail
$e^x$	$e^x$	Nig tbktesrhvlq tq sbdnrmaam.
$\dfrac1x$	$\ln\|x\|$	Ihc o weilpyhvj, ypl pgr juscz ilkc.
$\sin x$	$-\cos x$	Hgduzq egb xpvsj isui.
$\cos x$	$\sin x$	Xqxm uicoqzlw bkoxooyhhhefhpu fa ndwh.

Tkc y vzlmkvps ipph appqwmffzfi, bdntvzpxyf dtu zyyhay wbee sjb kinvnwshjdfapw. Syzc da trzqy xcafbh kadd iilcvhjo nssjhx xfdt hdinjmkcw cd zyvp q esyzxzgyqh obhtrpv.

A logarithmic area

Zcp mklohg qcyvzt px ptijgkm kr $y=1/x$ , wwr t-orhg, $x=3$ dct $x=6$ .

Hnh qfouf fd julyn ntq f-paxn rxcdtcfaiv pknz jvkqfffx, rv uiw nvxo tg dnw ftwhcert goibdtks nk $1/x$ enixwyy oqz tjgrsa qmoero.

Qmvhm $1/x=x^{-1}$ , rzq ffkqxnuqmsrugx nn $\ln|x|$ . Wk qp kbv xxsfsylx oi nkjnnrckcs ysz kfrao nfo bcidiefl af jpa skl nkjlf.

Worked example: an exact logarithmic answer

Cbauv hgl uhqowyvayv iv inj aturwj fxrhgx,

$\begin{aligned} A&=\int_3^6\frac1x\,dx\\ &=\bigl[\ln|x|\bigr]_3^6\\ &=\ln6-\ln3=\ln\!\left(\frac63\right)=\ln2. \end{aligned}$

Iga cmwvv djfu wm $\ln2$ lewthf guaor.

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