Integration - the Australian Mathematical Sciences Institute

{24} • IntegrationExample1 Find∫ 20(−x 2 − x + 2) d x.2 Find the area of the shaded region.3 If your velocity at time t is given byv(t) = −t 2 − t + 2, then what is yourdisplacement and your distance travelledbetween time t = 0 and t = 2?Solutionyy = –x2 – x + 20 2x1 ∫ 20[(−x 2 − x + 2) d x = − 1 3 x3 − 1 ] 22 x2 + 2x0=(− 8 )3 − 2 + 4 − ( 0 ) = − 2 3 .2 The area is partly above and partly below the x-axis. Factorising−x 2 − x + 2 = −(x − 1)(x + 2),we find the x-intercept is at x = 1. Therefore the total area shown is∫ 10(−x 2 − x + 2) d x −∫ 2[= − 1 3 x3 − 1 2 x2 + 2x1(−x 2 − x + 2) d x] 1 [− − 10 3 x3 − 1 ] 22 x2 + 2x1( (= − 1 3 − 1 )2 + 2 − ( 0 )) ( (− − 8 )3 − 2 + 4 −(− 1 3 − 1 ) )2 + 2( 7) (=6 − 0 − − 2 3 − 7 )6= 7 (6 − − 11 )= 3.63 Your displacement is ∫ 20 v(t) d t, which was computed in part 1, and is − 2 3(i.e., youend up 2 3backwards of where you started). Your total distance travelled is given bythe area calculated in part 2, which is 3.Beware! Computing integrals often involves many fractions, subtractions and negativesigns, as in the above example! It is good practice to carefully bracket all terms, as shown.