5.2 The Definite Integral
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1 5.2 THE DEFINITE INTEGRAL 5.2 The Definite Integrl In the previous section, we sw how to pproximte totl chnge given the rte of chnge. In this section we see how to mke the pproximtion more ccurte. Suppose tht we wnt to find the totl distnce trveled over the time intervl t b. We tke mesurements of the velocity v(t) t eqully spced times, = t < t < t 2 < < t n = b. This mens tht we divide the intervl [, b] into n equl pieces ech of length t = b. We first use the left-end point n of ech intervl [t i, t i ] nd construct the left-hnd sum L(v, n) = v(t ) t + v(t ) t + + v(t n ) t. Geometriclly, this sum represents the sum of res of rectngles constructed by tking the height to be the vlue of the function t the left-endpoint of ech subintervl. See Figure Figure 5.2. Secondly, we use the right-end point of ech intervl [t i, t i ] nd construct the right-hnd sum R(v, n) = v(t ) t + v(t 2 ) t + + v(t n ) t. Geometriclly, this sum represents the sum of res of rectngles constructed by tking the height to be the vlue of the function t the right-endpoint of ech subintervl. See Figure
2 2 Figure Now, the exct distnce trveled lies between the two estimtes. As we hve seen in Section 5., by mking the time intervl smller nd smller we cn mke the difference between the two estimtes s smll s we like. This is equivlent to letting n. If the function v(t) is continuous then the following two limits re equl to the exct distnce trveled from t = to t = b. Totl distnce trveled = lim L(v, n) = lim R(v, n). n n Geometriclly, ech of the bove limit represents the re under the grph of v(t) bounded by the lines t =, t = b nd the horizontl xis. See Figure Figure Remrk 5.2. Notice tht for n incresing function the left-hnd sum is n underestimte wheres the right-hnd sum is n overestimte. This role is reversed for decresing function.
3 5.2 THE DEFINITE INTEGRAL 3 The bove discussion pplies to ny continuous function f on closed intervl [, b]. We strt by dividing the intervl [, b] into n subintervls ech of length t = b n. Let = t < t < < t n < t n = b be the endpoints of the subdivisions. We construct the left-hnd sum or the left Riemnn sum n L(f, n) = f(t ) t + f(t ) t + + f(t n ) t = f(t i ) t nd the right-hnd sum or the right Riemnn sum i= R(f, n) = f(t ) t + f(t 2 ) t + + f(t n ) t = n f(t i ) t. i= It is shown in dvnced clculus tht for continuous function on closed intervl [, b] tht s n both the left-hnd sum nd the right-hnd sum exist nd re equl. We denote the common vlue by the nottion b f(t)dt. b f(t)dt = lim n L(f, n) = lim n R(f, n). We cll b f(t)dt the definite integrl of f from t = to t = b. We cll the lower limit nd b the upper limit. The function f is clled the integrnd. Exmple 5.2. () On sketch of y = ln t, represent the left Riemnn sum with n = 2 pproximting 2 ln tdt. Write out the terms in the sum, but do not evlute. (b) On nother sketch of y = ln t, represent the right Riemnn sum with n = 2 pproximting 2 ln tdt. Write out the terms in the sum, but do not evlute. (c) Which sum is n underestimte? Which sum is n overestimte?
4 4 () The left Riemnn sum is the sum L(ln t, 2) = ln (.5) + ln (.5)(.5) =.5 ln (.5). The sum is represented by the rectngle shded to the left of Figure Figure (b) The right Riemnn sum is the sum R(ln t, 2) = ln (.5)(.5) + ln 2(.5) =.5 ln (2)(.5) =.5 ln 3. The sum is represented by the rectngles shded to the right of Figure (c) L(ln t, 2) < 2 ln tdt < R(ln t, 2) Definite integrls re used to find res. Tht is, definite integrl is the re under the grph of function. We will discuss this concept in the next section. Exmple (Estimting Definite Integrl from Tble) Use the tble to estimte 4 f(t)dt. Wht vlues of n nd t did you use? t f(t) The vlues of f(t) re spces units prt so tht t = nd n = b t =
5 5.2 THE DEFINITE INTEGRAL 5 4 = 4. Clculting the left-hnd sum nd right-hnd sum to obtin L(f, 4) = = 6, 45 R(f, 4) = = 7, f(t)dt 6, , 55 2 = 7, Exmple (Estimting Definite Integrl from Grph) The grph of f(t) is given in Figure Estimte 28 f(t)dt. Figure We pproximte the integrl using left nd right-hnd sums with n = 7 nd t = 4. L(f, 7) = = 56.6 R(f, 7) = = f(t)dt = 63.
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