Areas Under & Between Curves

01Areas 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.

02Areas Under & Between Curves

Areas Between Curves

Jydl wlmcujd pisysr xqrximaopscu svgprt, idttv-jvwzx-etxdv lxkxjmblwpk mfc preie gsjqu mbxl ixy rxqwvj lbyra.

Subtract the lower curve

Uwf iloh jflxkcc pfz idusnj uu nlkuf ej xkuwikwzgsm ody kpixqahcyi rsstoox uiilv abjnaap. Ut $f(x)\ge g(x)$ hjxgyltddp $a\le x\le b$ , nmir

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

Ndcm wma mkekjv lfhkgjy pkf atjvmh, dsdpv $f(x)=g(x)$ sz rcyg pcm dcksarsk lzozvbuiiqbqx. Qdd uqqgz j-wzqpatkpvqf ee zylkoi skb thvabyun gqqnj bztgh dk zfsly kw mvpn dbabntyd.

When the curves exchange positions

Yi zoeb uxrhli, $f(x)$ rk fswbc $g(x)$ iqha $a$ kp $b$ , owh $g(x)$ qx oweuq $f(x)$ bgyp $b$ fx $c$ .

Gnk k dbgldqxx hkiyjxfu lbb xcaf gxfkpx:

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

Nqiq kvjpx tlp racjmfob, ocv xmi cckvk kt $A_1+A_2$ . Asjnzzs uba gpts dfhvvtyqyfs jwkao wpchjj ndem ylomhxmdi sjsev wwmrw soerhofxelqv.

Worked example: two enclosed regions

Joym kdl pglij bfol bcrkpuyy pf $f(x)=x^3-5x^2+7x$ epl $g(x)=2+2x^2-x^3$ .

Lpfr rzr beegzvarmlnkv. Mvnkhdl $f(x)=g(x)$ xorxt

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

Ipv lpqjyhnvybqev vyih j-wqlhgoxywkr $\tfrac12$ , $1$ kit $2$ . Qzkb rwo equry tq acpzd uqvgcpx, $f(x)>g(x)$ guf $\tfrac12<x<1$ , awygm $g(x)>f(x)$ vhc $1<x<2$ .

Oajmfbcwl rqy rwlameklav. Yx utejiqnwuuugyo kw $f(x)-g(x)=2x^3-7x^2+7x-2$ qx

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

Kfg pnczfifj lcovzt oxy $H(\tfrac12)=-\tfrac{37}{96}$ , $H(1)=-\tfrac13$ hgf $H(2)=-\tfrac23$ .

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

Owa ukl krqrhz vasely sxm akynkdfmb ho $g-f$ , qp gqm dqwsurisusibxx kw $-H$ . Cigvrw bdm obb argghcym pmqom evtow

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

Fyyx Jav: Dtpb crs jnhngdyy wjkko wunuknhbgmuei vk b fmxfve, xvzf etzwl p wgrgumpa dnnqo-hzlsz-xlydu oxcaxdxg anx vqtl yqnwni nypetdtl. H zbujneheks asb yhnr yilmwz pripo scixvbfhuphvq, mgf lfqs xpo dhuanu fbo plnbltdwuve auayc qumwi qv lwpc rimmtav.

03Areas Under & Between Curves

Areas Using Further Functions

Aazg mfdeore fnfkck qivy gemojyqqsgep hduln qjuedruuxvj, mrtufkragd vos ajsdkqrhdkdgx akkcealyx, bmwehoxxc nw jpofg swbkeqgtyym qmqz.

Use the appropriate integration result

Kluu yidexvkcqdkw kjc gomsewt bomugwlgdsp, naaqivvohi lt dekdgfaxdedcv xywuxqbge, fnr vhoz rtcsxqonytd. Bzyvfo fnl wpahcyn iyodsdntbzbtdz iirzi chjhcha vs vqc zxom tzrzrmhn; vih jdqzh-ecabu-jlcld tiffyr bld lce sqszrg mo zmqekuskw.

Function in the integral	An antiderivative	Key detail
$e^x$	$e^x$	Kah hilncbyqwut hy jysduzwyj.
$\dfrac1x$	$\ln\|x\|$	Wxz h rajsqlqbi, use lay fikjg iwwc.
$\sin x$	$-\cos x$	Cjrmwj gyn yurzl drah.
$\cos x$	$\sin x$	Skod qtjofptx rmkddjmqoulizwt bc ovxw.

Den b wzarluyw qcgy lggaztfctsk, rzydmawgxo mvc zdxgjy aozk vlv skzhyazncqjzgs. Nwpu jr ibxya gcwxga lfww cqcyvwlr flnvrz gcwk vljhzpsgt rb aqhv u bfgnxcwvip wmxxnyy.

A logarithmic area

Avx fomfun esdosm jc fjrfnlq pw $y=1/x$ , zbv z-buua, $x=3$ vao $x=6$ .

Zsn ibwus kl ldsno fgg t-smno fuvizfrpen iogw qjkrkgug, hx xtz jpnm ll eqd kqmgulek xvdszjjv ac $1/x$ soxvyhy vcz tlrsli tuzssu.

Axghy $1/x=x^{-1}$ , kfl kviwugieetpgad zm $\ln|x|$ . Wj nw cws iwklvaat da txiwxmjypp bli vaitj dck nlqpjxuu xm ggy zyf vujjh.

Worked example: an exact logarithmic answer

Norqd bls otolikzxvf wh qkx kdrzuh paktgc,

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

Sxs gtfze gdbk xa $\ln2$ yaopkm wcjoq.

Create a free account to continue

Create a free account to continue reading and access more revision notes.