If f is a function defined for a x b, we divide the interval [a, b] into n subintervals of equal width x = b a n

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1 4.2 The Definite Integrl Recll: A limit of the form rises when we compute n re nd when we try to find the distnce trveled by n object. This sme type of limit occurs in wide vriety of situtions, even when f is not necessrily positive function. Thus, we will give limits of this form nme nd nottion. If f is function defined for x b, we divide the intervl [, b] into n subintervls of equl width x = b n. We let x =, x 1, x 2,..., x n = b be the endpoints of these subintervls nd we let x 1, x 2,..., x n be ny in these subintervls, so x i lies in the ith subintervl [x i 1, x i ]. Then the is provided tht this limit exists nd gives the sme vlue for ll possible choices of smple points. If it does exist, we sy tht f is on [, b]. The precise mening of the limit tht defines the integrl is s follows: For every number ε > there is n integer N such tht for every integer n > N nd for every choice of x i in [x i 1, x i ]. The symbol is clled n. It is n elongted S. In the nottion b f(x) dx, f(x) is clled the nd nd b re clled the ; is the nd b is the. For now, the symbol dx hs no mening by itself; b f(x) dx is ll one symbol. The dx indictes tht the independent vrible is x. The procedure of clculting n integrl is clled The definite integrl b f(x) dx is NUMBER. Thus, it does not depend on x. We could use ny letter in plce of x without chnging the vlue of the integrl. The sum is clled. So the definite integrl of n integrble function cn be pproximted to within ny desired degree of ccurcy by Riemnn sum. We know tht if f hppens to be positive, then the Riemnn sum cn be interpreted s sum of res of pproximting rectngles. Thus, the definite integrl b f(x) dx cn be interpreted s the re under the curve y = f(x) from to b if f(x) on [, b]. 1

2 If f tkes on both positive nd negtive vlues, then the Riemnn sum is the sum of the res of the rectngles tht lie bove the x-xis nd the negtives of the res of the rectngles tht lie below the x-xis. A definite integrl cn be interpreted s : where A 1 is the re of the region bove the x-xis nd below the grph of f, nd A 2 is the re of the region below the x-xis nd bove the grph of f. There re situtions in which it is dvntgeous to work with subintervls of unequl width. If the subintervl widths re x 1, x 2,..., x n, we hve to ensure tht ll these widths pproch in the limiting process. This hppens if the lrgest width, mx x i, pproches. So in this cse, the definite integrl becomes Not ll functions re integrble. However, the most commonly occurring functions re integrble. If f is continuous on [, b], or if f hs only finite number of jump discontinuities, then f is integrble on [, b]; tht is, the definite integrl b f(x) dx exists. If f is integrble, then the vlue of the integrl does not depend on the the choices of smple points. To simplify the clcultion of the integrl, we often tke the smple points to be right endpoints. Then x i = x i nd the definition of the integrl simplifies s follows. If f is integrble on [, b] then where x = b n nd x i = + i x. 2

3 Lter, it will be importnt to recognize limits of sums s integrls. When we use limit to evlute definite integrl, we need to know how to work with sums. The following formuls will be useful: (1) (2) (3) i = i 2 = i 3 = (4) (5) (6) c = c i = ( i ± b i ) = Exmple 1. Use the form of the definition to evlute 4 1 (x 2 4x + 2) dx. 3

4 Exmple 2. Express s n integrl on the intervl [π, 2π]. lim n ( ) cos xi x x i Exmple 3. ) Evlute the Riemnn sum for f(x) = 4 x 2, tking the smple points to be right endpoints nd = 2, b = 2, nd n = 4. b) Evlute x2 dx. 4

5 Exmple 4. The grph of g is shown. Estimte 4 g(x) dx with six subintervls using left endpoints. 2 Exmple 5. The tble gives the vlues of function obtined from n experiment. x f(x) ) Use the tble to estimte 9 f(x) dx using three equl subintervls with right endpoints. 3 b) If the function is known to be n incresing function, cn you sy whether your estimtes re less thn or greter thn the exct vlue of the integrl? 5

6 Note: We will lern n esier method for the evlution of integrls in the coming sections so tht we cn evlute something like the previous exmple. Exmple 6. Evlute the following integrls by interpreting ech in terms of re. ) 2 4 x2 dx b) 4 2 ( ) 1 2 x + 4 dx 6

7 Exmple 7. The grph of g is shown. Evlute ech integrl by interpreting it in terms of res. ) 2 g(x) dx b) 6 2 g(x) dx c) 7 g(x) dx 7

8 We often choose the ith smple point to be the right endpoint of the ith subintervl out of convenience. However, when pproximting integrls, it is usully better to let the ith smple point be the midpoint of the ith subintervl, which we denote by x i. Using the Riemnn sum with midpoints gives the following pproximtion. Midpoint Rule: where nd Exmple 8. Use the Midpoint Rule with n = 4 to pproximte π/2 cos 4 x dx to four deciml plces. 8

9 Properties of the Definite Integrl: b b f(x) dx = f(x) dx = c dx = b b c [f(x) ± g(x)] dx = cf(x) dx = f(x) dx + b c f(x) dx = Exmple 9. If 5 1 f(x) dx = 12 nd 5 4 f(x) dx = 3.6, find 4 1 f(x) dx. 9

10 Exmple 1. Find 5 f(x) dx if f(x) = { 2 x < 3 x x 3 Comprison Properties of the Integrl: If b, then 1. If f(x) for x b, then. 2. If f(x) g(x) for x b, then. 3. If m f(x) M for x b, then Proofs: 1

11 Exmple 11. Use the third comprison property to estimte 2 (x 3 3x + 3) dx. 11