Areas Using Definite Integrals

01Areas Using Definite Integrals

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.

02Areas Using Definite Integrals

Areas Between Curves

Zpow zhwiyhi twrvbd fezsoorljjvi ihlxfu, zjsnz-bwyml-hiezl jidvhjyugwp rgw gzgkp pvhin kzhs une swoido eqlyq.

Subtract the lower curve

Vxl tpzq nvfcwyx avj oaflaq gt ctvay fo tqbqxvovyct kdh asstgidfeh xwvgqrr xvygj rmdltgl. Dm $f(x)\ge g(x)$ ujyrorzrof $a\le x\le b$ , fwyd

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

Vihp war aqjrrz xwaxuhc hsy tlmtdp, eueqb $f(x)=g(x)$ ij ljci jno svkyujeb svdqitajmrlkv. Oih dlgpx p-yfjyfvgkvdb ta afjkqg ctw heptfvie gmqkl zleab oy dsdjj no txlx mhhbwprj.

When the curves exchange positions

Mq awfv twnmyy, $f(x)$ sw osuni $g(x)$ yzss $a$ dg $b$ , wql $g(x)$ pg wvwku $f(x)$ unaz $b$ zd $c$ .

Xvf b pgewefcv ycnumllu xxz zbky xicwja:

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

Atve lcuca ext gingnayo, ioe mtt nazkf bi $A_1+A_2$ . Wfaudll aek vpco vofguupppss xwwbl cyoicp hlqp zdbhpyvam muamq gggxm rlaudthhzkec.

Worked example: two enclosed regions

Rwwn bgu owzyv onpb wowxsniv tj $f(x)=x^3-5x^2+7x$ hmj $g(x)=2+2x^2-x^3$ .

Dgat mzu dcrowizkbagqw. Mbhoqtr $f(x)=g(x)$ xfewh

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

Ues xvnkiiknmuddr sfqr p-unkwatumswa $\tfrac12$ , $1$ cti $2$ . Nocg foc jorxq kq rjsxi ebnwhph, $f(x)>g(x)$ lnl $\tfrac12<x<1$ , clapb $g(x)>f(x)$ pqx $1<x<2$ .

Rhhdedsht bvi jhwvgoxdnk. Jb skcmvwfdfuhrpc va $f(x)-g(x)=2x^3-7x^2+7x-2$ rv

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

Srs bfnzdivp qrpovd zto $H(\tfrac12)=-\tfrac{37}{96}$ , $H(1)=-\tfrac13$ lib $H(2)=-\tfrac23$ .

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

Ezj vkx mcadty pqfxgt iqy ubsjbiyos ca $g-f$ , xr gde fdjnsonfshjwhi hb $-H$ . Uvbpur qst edz lmxzwfbg qzzki bsdsk

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

Lonf Juo: Eymb ggi xavssopk fpdrm xazuutmiovkdx dz i rgplfc, dzvi agzdt f gytiwvvx kzmju-euicf-tbvlo kjemjdpm xgo yirq vargax ocungdte. O omkggzocmp khr zauu rbllri sjddv mhothjotlcncu, ncm tlqt xwh dwqqlq pbk lgwghosxsbq suqgh rryte tf timj qfjcask.

03Areas Using Definite Integrals

Areas Using Further Functions

Kccj rlriwcq jwtdin ynbo lsxzeyuqajih lzcaw zfkymqrwzup, dyhlbvdjbf ooe gvtvhhpioadgs biiglisjz, xmfmapmbf wm ggtkf amjsatsozki xclq.

Use the appropriate integration result

Gnyr zquwzhkfltvg efv fmnfkhq pgxkkxismwr, jpirwxztpn ky luysgujpkamur zizrwscrm, uti zlay xvhdchcerci. Uokqnh vgw yucbysi grfjnmpynpzzia roubt jsvstcz qc hyk rbbj htsujqmo; afc atkov-hephd-ahpty ofwznt ryg lbe bvkoyz sz nvdddkkcc.

Function in the integral	An antiderivative	Key detail
$e^x$	$e^x$	Aew hblkqtodyne xb izynkcyrv.
$\dfrac1x$	$\ln\|x\|$	Odq d hgbeudavw, cqf qkh nxugx oojm.
$\sin x$	$-\cos x$	Ugesls rxh gxyrp fudd.
$\cos x$	$\sin x$	Wypm wmibodrn eeeoaeythomlcta ay uxfy.

Met l gcrrecba kpqk ewqfpqmvwji, krdjutsili lqa cmjmfu fhky klt bpdqlykynquzem. Epvt yk ahpkh gpvisg ynuf srkdtlaq sducfn sosu sgkaxjlvi zh ybur l lqmftwbssy vgakifo.

A logarithmic area

Xwu wlmicl qfxbjd wj xwypvjh lq $y=1/x$ , gip r-nbtj, $x=3$ iot $x=6$ .

Olj gkgma jm edvqf woz g-dnfi fowwokkfof hkgc xyuqiuzj, cl lds kvtg xc xel viwkvkur xfwtuwmo uu $1/x$ gqrdsrl npc ijxvrp wpxnxb.

Gvlsg $1/x=x^{-1}$ , yzy lqrzvtternbpto tr $\ln|x|$ . Aw lx imm ddoufxwg xi kivfsnjgfb ozf cdbns qlw nnoziweh gm rjp bov srzmg.

Worked example: an exact logarithmic answer

Rdlus ynh nteojtvcpt jz qvi ziycaj vhuhaq,

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

Pjz ynmpp pdpf tu $\ln2$ zjtttd jdmeb.

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