Math 106 Fall 2014 Exam 2.1 October 31, ln(x) x 3 dx = 1. 2 x 2 ln(x) + = 1 2 x 2 ln(x) + 1. = 1 2 x 2 ln(x) 1 4 x 2 + C

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1 Math 6 Fall 4 Exam. October 3, 4. The following questions have to do with the integral (a) Evaluate dx. Use integration by parts (x 3 dx = ) ( dx = ) x3 x dx = x x () dx = x + x x dx = x + x 3 dx dx = x 4 x + C u = dv = x 3 You can also do parts using u and dv: du = v= to get x x x 3 dx = ( ) x x x dx = x 4 x + C

2 (b) Suppose your answer from part (a) is the integral dx is convergent or divergent. Assuming the answer from (a) is ln x 4x x, the Fundamental Theorem of Calculus says To determine t dx = limit has indeterminate form ln x 4x x. Explain how to use this answer to understand if ( ln x 4x x ) t = ln t 4t t ln() 3 ln t 4t dx we would then take the limit of t and requires L Hospital s Rule: as t goes to infinity. This ( ln t 4 t lim t t = lim 4) = lim t 4t t 4t 4 4t = =. Since the limit produces a real number, the improper integral dx is convergent. (c) Now suppose that you suspect that x3/ over some interval [a, ). x3/ Use this idea along with the Comparison Test to determine if a dx is convergent or divergent. Remember, your answer must include sentences! Because and dx is convergent by the P -Test (with p = 3/ > ) we know x3/ a x3/ that dx is convergent by the Comparison Test. a

3 . Consider the following integral. x x + 6 dx (a) Use a rationalizing substitution and related steps to convert the integral to a Case I or Case II Partial Fractions result. Your final answer for this part should be an integral of a rational function. DO NOT SOLVE FOR COEFFICIENTS AND DO NOT INTEGRATE! Let u = x +. Then u = x + 6 x = u 6 and u du = dx so x x + 6 dx = u u 6 u du = u u u 6 du = u (u 3)(u + ) du (b) Suppose your result from part (a) is Write (u )(u + ) = A u + B u + + du. Evaluate this new integral. (u )(u + ) C. Then clearing fractions gives (u + ) = A(u + ) + B(u )(u + ) + C(u ). Selecting values for u now allows us to solve for A, B, and C: u = : 8 = A( + ) A = u = : 8 = C( ) C = 4 Now select any other value of u and use the values of A and C to find B: u = : = ( + ) + B( )( + ) + 4( ) B 4 = B = So and (u )(u + ) = u u (u + ) (u )(u + ) du = u du u + du + 4 (u + ) du = ln u ln u + 4(u + ) + C

4 3. Consider the following integral. cos 3 (x) sin 4 (x) dx (a) Evaluate cos 3 (x) sin 4 (x) dx cos 3 (x) sin 4 (x) dx = cos (x) sin 4 (x) cos(x) dx = [ sin (x)] sin 4 (x) cos(x) dx Let u = sin(x) du = cos(x) dx [ sin (x)] sin 4 (x) cos(x) dx = [ u ]u 4 du = u 4 u 6 du = u u7 7 + C = [sin(x)] [sin(x)]7 7 + C (b) Suppose a pdf is given by c cos 3 (x) sin 4 (x) if x π f(x, y) = otherwise. where c is some constant number. Use properties of a pdf and your answer from part (a) to find the value for c. For any pdf f(x) we must have f(x) dx = ( π ) Since our functions is zero over the intervals (, ) and, this means = According to the answer above, Then π/ implies that c = 3 = 7.. π/ f(x) dx = π/ ( [sin(x)] cos 3 (x) sin 4 (x) dx = = π/ c cos 3 (x) sin 4 (x) dx ) π/ [sin(x)]7 7 c cos 3 (x) sin 4 (x) dx = c 3 = 7 = 3

5 4. The following questions have to do with the integral I = sin(x) cos(3x) dx (a) Demonstrate the results of one application of integration by parts applied to cos(3x) cos(x) dx. I = cos(3x) cos(x) dx = cos(3x)(sin(x)) dx = sin(x) cos(3x) sin(x)(cos(3x)) dx = sin(x) cos(3x) 3 sin(x) sin(3x) dx (b) Continue your work from part (a) and evaluate I = cos(3x) cos(x) dx. I = sin(x) cos(3x) 3 sin(x) sin(3x) dx = sin(x) cos(3x) 3 ( cos(x)) sin(3x) dx = sin(x) cos(3x) 3[ cos(x) sin(3x) cos(x)(sin(3x)) dx] = sin(x) cos(3x) 3[ cos(x) sin(3x) + 3 cos(x) cos(3x) dx] = sin(x) cos(3x) + 3 cos(x) sin(3x) 9 cos(x) cos(3x) dx] = sin(x) cos(3x) + 3 cos(x) sin(3x) 9I Then I = sin(x) cos(3x) 3 cos(x) sin(3x) 9I implies that I = sin(x) cos(3x) + 3 cos(x) sin(3x) or that I = [sin(x) cos(3x) + 3 cos(x) sin(3x)] + C.

6 . Consider the function f(x) = xe x (a) Find a relevant Taylor Polynomial that will allow you to evaluate the limit. We should use a = since the limit is as x. We will need at least P (x) since there is ax in the denominator. The appropriate derivatives and values are given in the chart below: n f (n) (x) f (n) () xe x xe x + e x = ( x)e x e x ( x)e x = (x )e x Assembling the pieces now gives P (x) = f () () + f () ()! (x ) + f () () (x ) = x! x = x x. (b) Use your Taylor Polynomial answer to evaluate the following limit: From part (a) we see that x xe x lim x x x xe x x (x x ) x lim x x = lim x x = lim x x = lim =. x