Finding areas by integration - Mathematics resources

1 Finding areas byintegrationmc-TY- areas -2009-1 integration can be used to calculate areas . In simple cases,the area is given by a single definiteintegral. But sometimes the integral gives a negative answer which is minus the area , and inmore complicated cases the correct answer can be obtained only by splitting the area into severalparts and adding or subtracting the appropriate order to master the techniques explained here it is vital that you undertake plenty of practiceexercises so that they become second reading this text, and/or viewing the video tutorial on this topic, you should be able to: find the area beween a curve, thex-axis, and two given ordinates; find the area between a curve and thex-axis, where the ordinates are given by the pointswhere the curve crosses the axis.

2 Find the area between two area between a curve and area between two way of Finding the area between two mathcentre 20091. IntroductionWe can obtain the area between a curve, thex-axis, and specific ordinates (that is, values ofx), by using integration . We know this from the units on integration as Summation, and onIntegration as the Reverse of Differentiation. In this unit we are going to look at how to applythis idea in a number of more complicated The area between a curve and thex-axisLet us begin by exploring the following question: Calculate the areas of the segments containedbetween thex-axis and the curvey=x(x 1)(x 2). In order to answer this question, it seems reasonable for us to draw a sketch of the curve, asthere is no mention of the ordinates, or values, ofx.

3 To make the sketch, we see first that thecurve crosses thex-axis wheny= 0, in other words whenx= 0,x= 1, andx= 2. Next,whenxis large and positive, we see thatyis also large and positive. Finally, whenxis large andnegative, we see thatyis also large and negative. So if we join up these features that we havefound on the graph, we can see that the curve looks like = x(x 1)(x 2)12 ABNow the areas required are obviously the areaAbetweenx= 0andx= 1, and the areaBbetweenx= 1andx= 2. But there is a marked difference between these two areas in terms oftheir position. The areaAis above thex-axis, whereas the areaBis below it. In previous unitswe have talked only about calculating areas using integration when the curve, and thus the area ,is above thex-axis.

4 Does the position of the curve make any difference to the area ?In this example, we shall play safe and calculate each area separately. We know that the areaAis given by the integral fromx= 0tox= 1of the curvey=x(x 1)(x 2) =x3 3x2+ 2x;thusA= 10ydx= 10(x3 3x2+ 2x)dx=[x44 3x33+2x22]10=[x44 x3+x2]10= [14 1 + 1] [04 0 + 0]= mathcentre 2009 area B should be given by a similar integral, except that now the limits of integration are fromx= 1tox= 2:B= 21ydx= 21(x3 3x2+ 2x)dx=[x44 3x33+2x22]21=[x44 x3+x2]21= [164 8 + 4] [14 1 + 1]= 0 14= the two integrals have the same magnitude, but areaAis above thex-axis and areaBisbelow thex-axis; and, as we see, the sign of the valueBis negative.

5 The actual value of thearea is+14, so why does our calculation give a negative answer?In the unit on integration as Summation, when we summed the small portions of area we wereevaluating y x. Now xwas defined as a small positive increment inx, butywas simply they-value at the ordinatex. Clearly thisy-value will be negative if the curve is below thex-axis,so in this case the quantityy xwill beminusthe value of the area . So in our example, we wereadding up a lot of values equal tominusthe area , as the curve is wholly below thex-axis betweenx= 1andx= 2. For this reason, the calculation gives a negative answer which is minus thevalue of the suppose for a moment that the question had asked us instead to find thetotalarea enclosedby the curve and thex-axis.

6 We know the answer to this question. The total area is the area ofA,14of a unit of area , added to actual value of the areaB, which is another14of a unit of the total area enclosed by the curve and thex-axis is12of a unit of area . But supposewe had decided to work this out by Finding the value of the integral betweenx= 0andx= 2,without drawing a sketch. What answer would we get? We would find area = 20ydx= 20(x3 3x2+ 2x)dx=[x44 3x33+2x22]20=[x44 x3+x2]20= [164 8 + 4] [04 0 + 0]= 0 0= mathcentre 2009So according to this calculation, the area enclosed by the curve and thex-axis is zero. But weknow that this is not the case, because we have a sketch to prove it.

7 Clearly what has happened isthat the signed values of the two areas have been added together in the process of integration ,and they have cancelled each other the value of the integral evaluated between two ordinates is not necessarily the value ofthe area between the curve, thex-axis and the two ordinates. So we must be very careful whencalculating areas to avoid this particular trap. The best way is always to draw a sketch of thecurve over the required range of values PointWhen calculating the area between a curve and thex-axis, you should carry out separate calcu-lations for the parts of the curve above the axis, and the parts of the curve below the axis.

8 Theintegral for a part of the curve below the axis gives minus thearea for that part. You may findit helpful to draw a sketch of the curve for the required rangeofx-values, in order to see howmany separate calculations will be Some examplesExampleFind the area between the curvey=x(x 3)and the ordinatesx= 0andx= we sety= 0we see thatx(x 3) = 0, and sox= 0orx= 3. Thus the curve cuts thex-axisatx= 0and atx= 3. Thex2term is positive, and so we know that the curve forms aU-shapeas shown = x(x 3) mathcentre 2009 From the graph, we can see that we need to calculate the areaAbetween the curve, thex-axisand the ordinatesx= 0andx= 3first, and that we should expect this integral to give a negativeanswer because the area is wholly below thex-axis:A= 30ydx= 30(x2 3x)dx=[x33 3x22]30= [273 3 92] [03 3 02]= [9 272] [0]= , we need to calculate the areaBbetween the curve, thex-axis, and the ordinatesx= 3andx= 5.

9 B= 53ydx= 53(x2 3x)dx=[x33 3x22]53= [1253 3 252] [273 3 92]= 4123 3712 9 + 1312= the total actual area is412+ 823= 1316units of the area bounded by the curvey=x2+x+4, thex-axis and the ordinatesx= 1andx= we sety= 0we obtain the quadratic equationx2+x+ 4 = 0, and for this quadraticb2 4ac= 1 16 = 15so that there are no real roots. This means that the curve doesnotcross thex-axis. Furthermore, the coefficient ofx2is positive and so the curve 0,y= 4and so the curve looks like = x2 + x + mathcentre 2009 The required areaAis entirely above thex-axis and so we can simply evaluate the integralbetween the required limits:A= 31ydx= 31(x2+x+ 4)dx=[x33+x22+ 4x]31= [273+92+ 12] [13+12+ 4]= 2512 456which equals2023units of Find the area enclosed by the given curve, thex-axis, and the given ordinates.

10 (a) The curvey=x, fromx= 1tox= 3.(b) The curvey=x2+ 3x, fromx= 1tox= 3(c) The curvey=x2 4fromx= 2tox= 2(d) The curvey=x x2fromx= 0tox= 22. Find the area contained by the curvey=x(x 1)(x+ 1)and Calculate the value of 1 1x(x 1)(x+ 1)dx .Compare your answer with that obtained in question 3, and explain what has Calculate the value of 60(4x x2)dx .Explain your The area between two curvesA similar technique to the one we have just used can also be employed to find the areas sandwichedbetween the area of the segment cut from the curvey=x(3 x)by the liney= mathcentre 2009 SolutionSketching both curves on the same axes, we can see by settingy= 0that the curvey=x(3 x)cuts thex-axis atx= 0andx= 3.