Class Note for MATH 124 at UA
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Date Created: 02/06/15
A Chapter 6 Constructing Antiderivatives Section 64 Second Fundamental Theorem of Calculus Chapter 7 Integration Section 71 Integration by Substitution Math 124 Section 023 Fall 2007 Instructor Paul Dostert p1 A 64 2nd FToC The second fundamental theorem of calculus is essentially a restatement of the original FToC The idea is that if we want a function of a that is the antiderivative of f then we can simply integrate f from some number a to 13 Second Fundamental Theorem of Calculus If f is a continuous function on an interval and if a is any number in that interval then the function F defined as follows is an antiderivative of f my 2 70 ft dt Note This is the first FToC but we allow F to be a function of 23 rather than just a number Think of this as an integral formula where an endpoint the a endpoint can change p2 A 64 2nd FToC The important use of the 2ml FToC is in taking derivatives of integrals Clearly d b E ftdt0 because the integral is simply a constant What is d w t dt d9 f Well since we are integrating ft we have 70 m dt my Fm where F isaan antiderivative of f Thus m m at M W we Simply put the derivative of the integral is the function itself p3 A 64 2nd FToC Note The previous statement applies only if the limit in the integral is the variable we are taking the derivative of For example d 902 E mm a me What does it equal Well we use the chain rule d 002 d 2 ga ftdt gum Fa 2 o Ex Evaluate the following d a E O b 1 dz 5 d 2 y2y 1dy sint dt 627 cost p4 A 64 2nd FToC Ex Write an expression for f with the given properties a f 2 oos23 f0 1 b In 32 1 18 2 c ez l arctanz 7 Z Ex Evaluate the following 2 t2 tdt a d3 0 d cosy b it H dt dy siny d 3 c In 332 d9 p5 A 71 Integration by Substitution Recall from the chain rule we have d sin 332 2 cos 32 d3 Then what is Icos 32 d1 We can think of 22 as an inside function and a as an outside function The derivative of the inside function is found in the integral To find Icos 6132 dx we can choose u 32 to be the inside function Then du 23 daj To rewrite the integral in terms of u we need a 23 dd term We write 1 cos 32 2 dx We now rewrite the integral in terms of u 1 1 1 icosudu isinquC39 ism 32 C39 p6 A 71 Integration by Substitution We can think of integration by substitution as a backwards chain rule where we are searching for the inner function from chain rule Ex Evaluate the following a I 32 13 dx b se32 26 d6 cost cos sint dt lt gt cl e sin 9 dm C e e 3 d3 1234 p7 A 71 Integration by Substitution Sometimes it is simply not possible to find the integral via substitution For example I3 sin 32 1 dm What would our options for substitution be Why don t these work What do we do Sometimes we may have a function that is it not obvious will work What do we do then We guess For example CCC 1daj Clearly this doesn t come directly from a chain rule derivative We still treat the problem the same We choose u 2 1 thus du daj We have f Cu 1du f u32 u12 du c 152 x 132 C p8 A 71 Integration by Substitution Ex Evaluate the following a 3 dx 332 b 33 113C5 daj Ex Evaluate the following definite integrals 4 3 d3 a 0 V12CE 772 2 b a smajdgji 7T2 1CC6 1 c t22t3dt O 12 arosin d2 0 Vl CC2 d
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