10. AREAS BETWEEN CURVES


 Douglas Rice
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1 . AREAS BETWEEN CURVES.. Ares etween curves So res ove the xxis re positive nd res elow re negtive, right? Wrong! We lied! Well, when you first lern out integrtion it s convenient fiction tht s true in sense. But now we ll level with you nd tell you the whole truth. The oundries for n re tht s given y n integrl re the grph of function, two verticl lines nd, up to now the fourth oundry hs een the xxis. We re now going to generlise tht so tht the fourth oundry is the grph of second function. You cn still tke the second function to e y = nd tht gives the xxis so you cn still hve the xxis s oundry if you like. But you cn hve curved ottom oundry too. We re going to develop formul for the re etween two curves. There will e top y nd ottom y. Top nd ottom refer to which hs the greter y vlues. Of course curves my cross over nd this mkes the prolem just little it more complicted, ut not much. y = f(x) (top y) y = g(x) (ottom y) The shded re is the re we wnt. Now f(x) dx gives the re under the top curve, right down to the xxis nd g(x) dx gives the re under the ottom curve, right down to the xxis. Clerly the re we wnt is the difference etween the two, tht is, f(x) dx g(x) dx. But this cn e written s [f(x) g(x)] dx. If f(x) g(x) for x, the re etween y = f(x) nd y = g(x) etween x = nd x = is [f(x) g(x)] dx
2 . To rememer this, just rememer the integrl s: top limit ottom limit (top y ottom y) Now when we hve just one curve y = f(x) which lies ove the xxis from x = to x = nd the ottom oundry of the re is the xxis, then the top y is y = f(x) nd the ottom y is y =. So the re under the grph is [f(x) ] dx) = f(x) dx, s efore. But if y = f(x) lies elow the xxis this ecomes the ottom oundry nd the xxis, y = ecomes the top oundry. The re ecomes, in this cse, [ f(x)] dx = f(x) dx. Previously we explined this minus sign y sying tht res elow the xxis re negtive. Now we re in position to set the record stright. Ares re never negtive, ut integrls my e. Where the curve is elow the xxis the re is positive ut the integrl is negtive. Tht s the right wy to look t it. Whenever you hve to work out res elow, ove or etween curves lwys set it up s prolem involving the re etween two curves. If you correctly identify which is the top y nd which is the ottom y you ll never go wrong. The integrl you set up will lwys come out s positive nd will give the required re. Integrls come out negtive, when deling with res, only if you hven t set things up properly. There s lot of sloppy thinking with this topic nd students re often tught d hits. Alright we did strt you off thinking tht res elow the xis re negtive ut t the time it would hve confused you to do otherwise. At lest we re now setting the story right. Some students think tht since res re positive ut integrls cn e negtive, ll you hve to do is to put solute vlue signs round everything. Even worse is the student who, on evluting n integrl to e, writes = =. Using solute vlue signs disply ignornce s to wht is relly going on nd cn led to wrong nswers if the curves cross over. So some guidelines to void such errors re: Never fudge the sign if the integrl comes out negtive. Never use solute vlue signs in connection with res. Exmple : Find the re etween y = x, y = 4 nd x =. y = 4 y = x Solution: Top y is y= 4 nd ottom y is y = x. The re is therefore [4 x ] dx = [ 4x (/)x ] = = 5.
3 .. Wht If Curves Cross? It s solutely importnt to determine whether curves cross within the region eing considered. For if they do, wht ws the top y will ecome the ottom y nd vice vers. The intervl of integrtion will hve to e split up. Suppose we wnt the re etween the curves y = f(x) nd g(x) etween x = nd x = nd suppose the curves cross t some point c etween nd. Suppose tht g(x) f(x) for x c nd g(x) f(x) for x c. Then from x = to x = c it s g(x) which is the top y, while from x = c to x = it s f(x) tht s on top. We must rek the intervl [, ] into two pieces nd integrte seprtely over these two regions. y = g(x) y = f(x) The required re will e c c [g(x) f(x)] dx + [f(x) g(x)] dx. c Exmple : Find the re etween y = x, the xxis nd lines x = nd x =. Solution: Becuse the curve cuts the xxis t x = we must divide the intervl [, ] into two pieces. From x = to x = the top y is y = (the xxis) while from x = to x = it is y = x. The required re is [ x ] dx + [x ] dx = x4 4 + x 4 4 = =. Two things should e noted with this exmple. () If we hd simply integrted x etween nd we would hve concluded tht the re is x dx = x 4 4 = 4 4 =, which is clerly wrong. And tking solute vlues wouldn t help either. Rememer to never use solute vlues with these re questions. If you don t know which is top y nd which is ottom you re likely to come up with totlly wrong nswer. If you do, then you ll set up the integrl correctly so tht it will come out positive. () We could hve exploited symmetry to simplify the clcultion nd void ll those minus signs. The re elow the xxis is clerly the sme s the re ove so we could hve sid tht the totl x 4 re = x dx = 4 =.
4 Exmple : Find the re etween y = x nd y = x. Solution: Here we re not given ny endpoints. Tht s ecuse the grphs of y = x nd y = x cut in exctly two plces nd so they enclose region. We need to find these two points. Solving x = x we get x = or. So these re the limits of integrtion. The top y, etween nd, is y = x. The re is therefore [x x ] dx. Becuse we know wht we re doing nd hve set up the integrl correctly we cn e sure tht it will e positive. x [x x ] dx = x = = 6. This is the required re. Exmple 4: Find the re etween y = x, y = x nd x =. Solution: The curves cut when x = x, tht is when x =. So they cut in two plces, t x =. / / We hve to integrte from to nd then from to. By symmetry the integrl from to is doule tht from to. / The re is therefore / = [( x ) x ] dx + [x ( x )] dx / [ x ] dx + [x ] dx = x / x / + x x = + = (6 + ) = 6 = =. / 4
5 EXERCISES FOR CHAPTER Exercise : Find the re etween the prols y = x, y = x nd the yxis. Give your nswer s n exct vlue, in terms of, s well s n pproximtion to 4 deciml plces. Exercise : Find the re enclosed etween the curves y = x x nd x x. Exercise : Find the re enclosed etween the curves y = e x nd y = 6x x etween x = nd x =. Give your nswer s n exct vlue involving e s well s n pproximte numericl vlue to 4 deciml plces. Exercise 4: Find the re enclosed etween the curve y = 6 x nd the line y = 5 x. Give the nswer to 4 deciml plces without using your clcultor, ut insted using the fct tht, to 4 deciml plces, log.5 =.455. Exercise 5: Use integrtion to find the re etween the lines y = + x nd y = x etween x = 5 nd x =. Check your nswer using simple geometry. Exercise 6: Find the re enclosed etween the curves y = 4x + 5, y = 5x + 4 nd the xxis. Exercise 7: The Lorenz curve is grph tht s used in economics tht shows the distriution of income. If the ottom x% of households represents the ottom y% of income, the point (x, y) lies on the Lorenz curve. The line y = x corresponds to perfect equlity, ut in prctice the curve lies elow it. The most extreme cse, (perfect inequlity) would e where one household erns ll of the income nd the rest ern nothing. This corresponds to the line y = (with spike t x = ). The Gini coefficient is the rtio of the re etween the line of perfect equlity nd the oserved Lorenz curve, nd the re etween the line of perfect equlity nd the line pf perfect inequlity. It is mesure of inequlity. The higher the Gini coefficient, the more unequl is the distriution of income. Find the Gini coefficient if the Lorenz curve is y = x. line of perfect equlity oserved Lorenz curve line of perfect inequlity 5
6 SOLUTIONS FOR CHAPTER Exercise : The curves cut when x = x, tht is, when x = or x =. Over the rnge from x = to x = the top y is y = x. / So the re is / ( x ) x dx = / x dx = [ x x ] =..7. Exercise : The curves cross if x x = x x. Solving, we get x x + x =. Fctorising we get x(x )(x ) =, so the curves cross when x =, nd. The re enclosed is the region etween the curves from x = to x =. But since they cross t x = we need to split the intervl [, ] into two pieces. From x = to x = the top curve is y = x x. Then from x = to x = the top curve is y = x x. So the re is: A = [(x x ) (x x)] dx + [(x x) (x x )] dx = x 4 4 x + x + x x4 4 x = =. Exercise : The curves do not cross nd the top y is y = e x. So the re is [e x (6x x )] dx = = (e ) = e [e x 6x + x + ] dx) = e x x + x +x] Exercise 4: The curves cut when 6 x = 5 x. Solving we get 6 = 5x x nd so x 5x + 6 =. Hence (x )(x ) = nd so x =, 5. When x =.5 the vlue of y for the curve y = 6 x is.4 while for the line y = 5 x it is.5. So etween x = nd x = the top y is y = 5 x. 6
7 The re etween is therefore [(5 x) x ] dx = [ 5x x log x ] = (5 9 log ) (9 4 log ) = log + log = log(/).455 = Exercise 5: The lines cut t x =. When x =, the vlue of y for y = + x is while for y = x it is, so etween x = 5 nd x = the top y is y = x. Between x = nd x = the top y is y = x +. So the enclosed re is 5 [( x) (x + )] dx + [(x + ) ( x)] dx = [ x] dx + [x] dx = x x = ( +.5) + ( ) = 75 + = 78 = 9. We cn check this y simple geometry since the region consists of two tringles. Rememer tht the re of tringle is ½ se perpendiculr height. Here we tke the se to e the verticl sides. When x = 5, x = 6 while x + = 9. So the lrger tringle hs se of 5 nd perpendiculr height of 5. Its re is 5 5 = 75. When x =, x + = nd x = so the smller tringle hs se nd perpendiculr height of. Its re is. The totl re is therefore 75 + = 9. Exercise 6: The curves cross when 4x + 5 = 5x + 4, tht is when x =. 7
8 The curve y = 5x + 4 cuts the xxis t x = 4/5 while y = 4x + 5 cuts it t x = 5/4. We must divide the re into two portions t x = 4/5. The top y etween x = 5/4 nd x = is 4x + 5. Between x = 5/4 nd 4/5 the ottom y is y =, the xxis. Then etween x = 4/5 nd x = the ottom y is y = 5x /5 Hence the re is 4x + 5 dx + [ 4x + 5 5x + 4] dx 4/5 5/4 4/5 =. 4 (4x +5)/ + 5/4. 4 (4x +5)/. 5 (5x + 4)/ = / / / = 6 / / 9 5 / 9/ = = = = 9 =.45. 4/5 [ 9/ ] Exercise 7: The re etween the line of perfect equlity (y = x) nd the oserved Lorenz curve, is (x x dx) = x x = = 6. The re etween the line of perfect equlity (y = x) nd the line of perfect inequlity (y = ) is the re of the tringle elow the line y = x, which is clerly ½. Hence the Gini coefficient is /6 / =. 8
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