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This 5 page Class Notes was uploaded by Jack Magann on Sunday March 1, 2015. The Class Notes belongs to MTH 162 at University of Miami taught by Dr. Bibby in Spring2015. Since its upload, it has received 83 views. For similar materials see Calculus 2 in Mathematics (M) at University of Miami.
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Date Created: 03/01/15
Calculus 2 Chapter 66A Improper Integrals Integrals that are over an in nite interval This means there is no nite area under the curve There are three cases for improper integrals l faoo f x dx Here the interval is a through all values in the 00 direction This can be solved by taking the limit of the integral For most functions the integral as the left most point of the region t approaches in nity is going to one set value that can be calculated by its limit This is the process to solve this is fa fxdx 11mm fa39fxdx limtooFt Fa Where F is the antiderivative of fX 2 If fxdx This can be thought of the same as the previous case but in the opposite direction This is solved by fb fxdx limtoo fbtxdx limtooFt Fa 3 If fxdx Here there is an endless interval in both directions f ofxdx ffoofxdx famfxdx SO f ofxdx limtooFt Fa limtooFt Fa If such an integral as these has a nite limit solvable limit the integral is said to converge at the value of the integral If such an integral as these has a limit that does not exist the integral is said to diverge Ex 1 flooxizdx setootot o 1 t1 1 1 2dx11m 2dx11m 11m 1011 1 t X39 tgtOo a x t gtoo x t gtoo oo1 limtoo flooidx lirg lnltl 1n1 limtoo 1nt 0 00 Since the limit of the integral approaches 00 it diverges 3 fooo xe x dx t limtoo xe39xdx limtoo xe39x 6quot 6 limt mo ex10 Use the tabular method to nd the antiderivative t limtoo 1 Since F t here is 3 use L Hopital s Rule 11mm 1 0 1 1 4f d quot 0 x21 00 dx dx hmt mo f0 x21 fx21 tan 1x limtootan 1t tan 10 limtootan 1t 0 TLquot Knowing the graph of tan 1x below there are horizontal asymptotes at Eand 2 So limtootan 1t O 2 Before the next example it is important to review how integrals were rotated around an axis to form a 3D shape and gives its volume Recall the disk method As seen in the picture the disc method takes the shape formed by rotating around either the X or yaXis makes thin slices of it and adds them up to calculate the volume This method then leads us to taking the integral of the area of the disks cylinders formed so A m 2 h rfx hAxdx Now that you remember that 5 Let R be the region bounded by the graphs of y i and y 0 XaXis a Find the area of R ex problem 2 This equation has an in nite area but that doesn t mean the solid it makes does too b Find the volume of the solid formed by revolving R around the XaXis V f1007ti2dx nflw 12 dx nflwx de x t t 1 t 1 1 l1mtoo nfl de l1mtoo nfl x de 11mtoo 7T 1 limtoo 7T 16 7T0 1 2 7T Not every integral that is divergent makes a solid that s convergent These special cases are referred to as Gabriel s Horn due to the solid created Calculus 2 Chapter 66B Discontinuous Integrands Limit approaches a vertical asymptote This means f x will be discontinuous at some point Recall A vertical asymptote is created Where the denominator equals 0 Case 1 Suppose x a is a vertical asymptote for f x Then ffxdx 11mm ft fxdx limtaFb Ft Note that if the limit exists then Ft exists This is integration from a vertical asymptote Case 2 Suppose x b is a vertical asymptote for f x Then ff fxdx 11mm f fxdx limtbFt Fa This is integration to a vertical asymptote Case 3 Suppose x c is a vertical asymptote for f x Where a lt c lt b b b Then fa f xdx f xdx fc f xdx 2 11mm f fxdx 11mm ft fxdx 102 Fa Fb Ft in case three for the integral to converge both limits must exists Ex 1 01 3 Use case 1 since there is a vertical asymptote at x 0 Therefore this is an example of integration from a vertical asymptote 1 dx 11Int gt0 ft TE llmt gt02 11Int gt02 2 2 2 f2 d x Use case 2 since there is a vertical asymptote at x 2 0 96 2 This is an example of integration to a vertical asymptote t d 11mt2 f0 96sz 11mt2lnx 205 11mt2lnlt 2 lnI 2 limt2ln0 ln2 00 This integral diverges 3 102 Vertical asymptote at x 2 limt2 fozx Ldfx limt2 sin391 limt2 sin391 sin391 0 lim sin 15 E t gt2 2 2
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