Section 6.1 Definite Integral

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1 Section 6.1 Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot e determined y known formul, so we ll need method for pproximting the re. Suppose we wnt to find the re under the prol nd ove the x xis, etween the lines x = 2 nd x = -2. We cn pproximte the re under the curve y sudividing the intervl [-2, 2] into smller intervls nd then drw rectngles extending from the x xis up to the curve. Suppose we divide the region into two prts nd drw two rectngles. We cn find the re of ech rectngle nd dd them together. Tht will give us n pproximtion of the re under the curve. This would not give very good pproximtion, s lrge region in Qudrnt 2 will e left out in the pproximtion of the re, nd lrge region in Qudrnt 1 will e included nd should not e. Section 6.1 Definite Integrl 1

2 Now suppose we increse the numer of rectngles tht we drw to four. We ll find the re of ech of the four rectngles nd dd them up. Here s the grph for this sitution. The pproximtion will e more ccurte, ut it still isn t perfect. Let s increse the numer of rectngles to 8: As we dd more nd more rectngles, the ccurcy improves. We re still not to n exct re, ut the re we d find using more rectngles is clerly more ccurte thn the re we d find if we just used 2 rectngles. Suppose we let the numer of rectngles increse without ound. If we do this, the width of ech rectngle ecomes smller nd smller, s the numer of rectngles pproches infinity, there will e no re tht is included tht shouldn t e nd none left out tht should e included Using left endpoints is not the only option we hve in working these prolems. We cn lso use right endpoints or midpoints. The first grph elow shows the region with eight rectngles, using right endpoints. The second grph elow shows the region with eight rectngles, using midpoints. Right Endpoints Midpoints Section 6.1 Definite Integrl 2

3 To get n exct re, we would need to let the numer of rectngles increse without ound: Alim f x1 f x2 f xn x n This lst computtion is quite difficult, we will not work prolem of this type. Insted, we will use limited numer of rectngles in the prolems tht we work. The process we re using to pproximte the re under the curve is clled finding Riemnn sum. These sums re nmed fter the Germn mthemticin who developed them. Approximting the re under curve given the type of Riemnn sums 1. Strt y finding the width of ech rectngle. A prtition of closed intervl, is finite suset of, tht contins the points nd. The lengths of these suintervls my or my not e equl. If the lengths re equl, it is clled regulr prtition nd x. n 2. Now find the height of the rectngles. Use the pproprite point in ech suintervl to compute the vlue of the function t ech of these points (gives the heights of the rectngles). 3. Find the re of ech rectngle nd dd them up. S* P f x1x 1 f x2x 2... f xn xn Section 6.1 Definite Integrl 3

4 Exmple 1: For ech prolem, pproximte the re under the curve over the given intervl, with the given numer of prtitions nd type of Riemnn sums. 1 x. Given f x, use left endpoints from 1, 2 with n = 4. Now try it gin, ut use the right endpoints of ech sudivision. Section 6.1 Definite Integrl 4

5 2. Given fx 0.5x, use midpoints from 0,3 with n = 3. Section 6.1 Definite Integrl 5

6 We cn lso pproximte this re y using Upper Sums or Lower Sums. Upper Sums nd Lower Sums Let f e continuous function on, nd P x x x e prtition of,,,..., n 0 1 The upper sum of f is f n n. U P M x M x M x M x. The vlue M i is the mximum vlue of the function for prtition. L P mx m x m x m x. The vlue m i is the The lower sum of f is minimum vlue of the function for prtition. f n n Exmple 2: Find the upper sum for f (x) = 1 - x 2, x [ 1, 1] if the prtition is 3 1 P 1,,, Section 6.1 Definite Integrl 6

7 Try this one: Find the lower sum for f (x) = x 2, x [ 1, 1] if the prtition is P 1,,,, Keep in mind tht the mx or min does not hve to hppen t n endpoint of sudivision. You ll need to grph the originl function to figure this out. Section 6.1 Definite Integrl 7

8 Exmple 3: Find Lf ( P ) given f ( x) sinx over 0, nd 2 P 0,,, 4 3. Section 6.1 Definite Integrl 8

9 As the numer of prtitions re dded, the upper sum tends to get smller. As the numer of prtitions re dded, the lower sum tends to get igger. The numer they meet t is clled the definite integrl. For function f which is continuous on,, there is one nd only one numer tht stisfies the inequlity, for ll prtitions P of L f P I U f P,. This unique numer I is clled the definite integrl (or just the integrl) of f from to nd is denoted y f () xdx. We red f () xdx s: the integrl from to of f with respect to x. The component prts hve these nmes: : the integrl sign : lower limit of integrtion : upper limit of integrtion f x : integrnd dx indictes the independent vrile in discussion nd denotes the widths re getting smller. The procedure of clculting the integrl is clled integrtion. In generl, the integrl cn e negtive, positive or zero. 6 2 Try this one: Estimte 3x dx y using left endpoint estimtes, where n = 6. 0 Section 6.1 Definite Integrl 9

10 Importnt Properties of Definite Integrl Assume tht f nd g re continuous functions. 1. f () x dx 0 2. f () xdx f() xdx When we defined the definite integrl () f xdx. However, the integrl mkes sense even if integrte from right to left., we ssumed tht. We cn f x g x dx f x dx g x dx kf x dx k f x dx, where k is constnt numer. Section 6.1 Definite Integrl 10

11 Exmple 4: Given 1 the following integrls f x dx 5 5 f x dx10, gx dx 4, f x dx 6, f x dx 8. Evlute f x g x dx 6 c. 5 f x dx 1 d. gx 5 dx Try this one: Given f( x) dx 3, f( x) dx 5, f( x) dx 9, find f( x) dx Section 6.1 Definite Integrl 11

12 Are Under the Grph of Nonnegtive Function If y f x is nonnegtive nd integrle over the intervl,, then the re under the curve y f x over, is given y f xdx 0. If the curve is sometimes negtive, then one cn split the region into pieces using the roots of the function s the limits of the integrl. Consider the function whose grph is given elow: Theorem: If f is continuous on, nd if c For the function shown ove, Are of f 1 ( xdx ) nd c. Are of 2 f ( xdx ) f( xdx ) c c c f x dx f x dx f x dx, then Section 6.1 Definite Integrl 12

13 Exmple 5: Given the grph of f, if the re of 1 is 12 nd re of 2 is 8, find c f xdx. Exmple 6: Given 1 f x dx 5, f x dx 12, 1 1 the curve nd the x-xis from x = -1 to x = f x dx 4. Find the re etween Section 6.1 Definite Integrl 13

14 Other Properties of Definite Integrl Assume tht f nd g re continuous functions., where k is constnt numer. 1. kdx k 2. If f x gx over,, then f x dx g x dx. 3. If m f x M over,, then m f xdx M f x dx f x dx If f is n odd function, then f xdx 0.. If f is n even function, then 2 f x dx f x dx. 0 Section 6.1 Definite Integrl 14

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