Math 132 Fall 2007 Exam I Notation D f x means the derivative of f evaluated at x For example if 3 2 f x x then D f x 3 x and D f 5 75 N f s x for a function f on an interval a b is said to be an 1 A Riemann sum j 1 j upper Riemann sum if for each j th the point s in the j subinterval is chosen so that f s is the maximum value of f x for j j th x in the j subinterval Using the uniform partition of a b 0 6 into 3 subintervals calculate the upper Riemann sum for the function f whose graph is given a 13 f 18 b 14 g 19 Solution h c 15 h 20 d 16 i 21 e 17 j 22 Delta 6 0 3 upperRiemannSum f 0 5 Delta f 2 Delta f 2 Delta upperRiemannSum subs f 0 5 4 f 2 3 upperRiemannSum 2 upperRiemannSum 2 f 0 5 4 f 2 upperRiemannSum 20 2 The expression is a right Riemann sum What is lim SN N a 16 f 44 b 20 g 56 Solution h c 24 h 60 d 32 i 64 e 36 j 72 A right Riemann sum of f x on the interval a b has the form restart f a b a N b a N f a 2 b a N b a N f a N b a N b a N b a b a f a N 2 b a b a f a N N N f b b a N Comparing this with the given sum we see that a 2 Delta 2 N b a N Delta f x x 3 a 2 2 N b 4 f x x3 Since the limit of the Riemann sums is the Riemann integral of f over a b we have Answer Int f x x a b answer value Answer 4 Answer x3 dx 2 answer 60 3 2 3 Calculate sec d 0 a 1 b 2 c 3 d 2 1 e 3 1 f 3 2 g 2 2 h 2 3 i 3 2 j 3 3 Solution c indefiniteIntegral Int sec theta 2 theta indefiniteIntegral sec 2 d indefiniteIntegralEvaluated value indefiniteIntegral indefiniteIntegralEvaluated sin cos answer subs theta Pi 3 indefiniteIntegralEvaluated subs theta 0 indefiniteIntegralEvaluated answer sin 3 sin 0 cos 0 cos 3 simplifiedAnswer simplify answer simplifiedAnswer 3 4 Suppose that f x a 9 f 1 b 7 g 3 4 x2 4 x 20 x2 x c 5 h 5 4 What is D f x dx 1 d 3 i 7 e 1 j 9 Solution a restart f x 4 x 2 4 x 20 x 2 x f x answer f 4 f 1 I 4 x2 4 x 20 x2 x Fundamental Theorem of Calculus Part answer 9 4 1 4 9 5 Suppose that 2 f x dx 5 f x dx and c f x dx 22 2 1 1 1 a 1 f 6 b 2 g 7 c 3 h 8 What is c e 5 j 10 d 4 i 9 Solution c restart 4 eqn1 c f x dx 22 1 4 eqn1 c f x dx 22 1 eqn2 c solve eqn1 c 4 1 22 eqn2 c f x dx 5 5 1 eqn3 subs int f x x 1 4 int f x x 1 1 1 2 int 2 f x x 1 4 eqn2 1 4 1 1 22 eqn3 c f x dx 2 f x dx 5 10 5 1 9 subs f x dx 2 f x dx 5 eqn3 2 1 1 c 3 1 4 1 2 6 Suppose that f x 3 x 2 x The Mean Value Theorem for Integrals asserts that there is a point c in the interval 1 4 such that f c f where f is the average value of f x ave ave for x in the interval 1 4 What is c a 4 3 f 3 b 5 3 g 10 3 c 2 h 11 3 d 7 3 i 5 2 e 8 3 j 7 2 Solution e a 1 b 4 f x 3 x 2 2 x a 1 b 4 f x 3 x2 2 x m Int f x x 1 4 b a 4 1 m 3 x2 2 x dx 3 1 m value m m 16 solve f c m c 8 3 Only the first solution is in the interval 2 7 Calculate a 0 f 6 x d 7 t 22 dt dx t3 4 0 b 1 g 9 c 2 h 12 d 3 i 16 at x 2 e 4 j 18 Solution d f t 7 t 22 t 3 4 f t 7 t 22 t3 4 Answer subs t 2 f t Answer 3 8 Suppose that F x 0 a 16 f 44 b 20 g 48 x 2 2 t 20 dt c 24 h 60 What is D F 2 d 32 i 64 e 36 j 72 Solution c F x Int sqrt t 2 20 t 0 x 2 F x x 0 2 t2 20 dt D F x x4 20 2x D F 2 4 36 simplify D F 2 24 9 Suppose that F x sin x 1 2 t dt 2 What is D F 4 cos x a 0 b 1 2 f 4 g 2 c 1 d 2 h 2 2 i e 3 2 2 j 3 2 Solution g with student v x sin x u x cos x F x Int sqrt 1 2 t 2 t u x v x v sin u cos F x v x 1 2 t 2 dt u x eqn D F x subs t v x integrand F x Diff v x x subs t u x integrand F x Diff u x x d 1 d 2 4 sin x 2 sin x 2 4 cos x 2 cos x 2 dx 2 dx answer simplify subs x Pi 4 value rhs eqn eqn D F x 1 answer 2 4 x 1 3 10 Calculate dx x 0 a 4 f 24 b 8 g 28 c 12 h 32 d 16 i 36 e 20 j 40 Solution j restart Integral Int sqrt x 1 3 sqrt x x 0 4 4 3 x 1 Integral dx x 0 value Integral Answer 40 Here are the substitution steps with student substitution u sqrt x 1 substitution u x 1 equivalentIntegral changevar substitution Integral u equivalentIntegral 4 1 2 u 3 du 1 simplify value equivalentIntegral 40 3 11 Calculate 15 x x 2 dx 2 a 10 f 20 b 12 g 22 c 14 h 24 d 16 i 26 e 18 j 28 Solution i restart Integral Int 15 x sqrt x 2 x 2 3 3 Integral …
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