MAT126 Final 2009 Solutions Part I - MAT126 Fall 2009...

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MAT126 Fall 2009 Practice Final The actual Final exam will consist of twelve problems 1
Problem 1 1. Evaluate R π/ 2 π/ 3 sin 3 ( x ) cos 2 ( x ) dx 2. Evaluate R sin 2 ( x ) dx Solution: 1. First, recognize that the integrand consists of sin( x ) raised to an odd power, multiplied by cos( x ) raised to an even power. Such a function is a prime candidate for the substitution u = cos( x ) (so that one of the sin( x ) can be used for the change of variable, and the rest may be expressed in terms of cos( x ) via the identity sin 2 ( x ) = 1 - cos 2 ( x )). Therefore we substitute u = cos( x ), whereby du = - sin( x ) dx , and we can write (noting that u = cos( π/ 3) = 1 / 2 when x = π/ 3, and u = cos( π/ 2) = 0 when x = π/ 2) Z x = π/ 2 x = π/ 3 sin 3 ( x ) cos 2 ( x ) dx = Z x = π/ 2 x = π/ 3 ( - sin 2 ( x ) cos 2 ( x ))( - sin( x ) dx ) = Z x = π/ 2 x = π/ 3 ( - 1 + cos 2 ( x ))(cos 2 ( x ))( - sin( x ) dx ) = Z u = 0 u = 1 / 2 ( u 2 - 1)( u 2 ) du = Z u = 0 u = 1 / 2 ( u 4 - u 2 ) du = u 5 5 - u 3 3 u = 0 u = 1 / 2 = (0) 5 5 - (0) 3 3 ! - (1 / 2) 5 5 - (1 / 2) 3 3 ! = - 2 - 5 5 + 2 - 3 3 2. There are two ways to approach this problem. The first is to use the identity cos(2 x ) = cos 2 ( x ) - sin 2 ( x ) = 1 - 2 sin 2 ( x ) to get sin 2 ( x ) = 1 2 (1 - cos(2 x )) 2
This implies that Z sin 2 ( x ) dx = Z 1 2 (1 - cos(2 x )) dx = Z 1 2 dx - Z 1 2 cos(2 x ) dx = 1 2 x - 1 2 Z cos(2 x ) dx = 1 2 x - 1 2 sin(2 x ) 2 + C = 1 2 x - 1 4 sin(2 x ) + C The second approach is to use an integration by parts, setting u = sin x and dv = sin xdx ; this means that du = cos xdx and v = - cos x . Therefore, integration by parts gives Z sin 2 ( x ) dx = (sin x )( - cos x ) - Z ( - cos x )(cos xdx ) = - sin x cos x + Z cos 2 ( x ) dx Now we plug in the identity cos 2 ( x ) = 1 - sin 2 ( x ) to get Z sin 2 ( x ) dx = - sin x cos x + Z (1 - sin 2 ( x )) dx = - sin x cos x + Z dx - Z sin 2 ( x ) dx Adding R sin 2 ( x ) dx to both sides of the equation gives 2 Z sin 2 ( x ) dx = - sin( x ) cos( x ) + Z dx = - sin( x ) cos( x ) + x + C so that Z sin 2 ( x ) dx = 1 2 x - 1 2 sin( x ) cos( x ) + C Notice that both approaches give the same answer, because of the identity sin(2 x ) = 2 sin( x ) cos( x ). 3
Problem 2 1. Estimate the integral 8 Z 7 dx ln( x ) using three rectangles and (a) right endpoints (b) left endpoints (c) Are your answers in 1a and 1b over- or under-estimates of the actual inte- gral? 2. Do the same for the under-integral function f ( t ) = e t 3 Solution: 1. The problem asks for three rectangles, so we divide the interval [7 , 8] into three equal subintervals: " 7 , 7 1 3 # " 7 1 3 , 7 2 3 # " 7 2 3 , 8 # Each subinterval has length 1 3 . (a) For right endpoints, evaluate f ( x ) = 1 ln( x ) at the right endpoint of each subinter- val, and multiply the sum by 1 3 , the length of the subintervals: 1 3 " f 7 1 3 ! + f 7 2 3 ! + f (8) # = 1 3 1 ln 7 1 3 + 1 ln 7 2 3 + 1 ln(8) (b) For left endpoints, evaluate f ( x ) = 1 ln( x ) at the left endpoint of each subinterval, and multiply the sum by 1 3 , the length of the subintervals: 1 3 " f (7) + f 7 1 3 ! + f 7 2 3 !# = 1 3 1 ln(7) + 1 ln 7 1 3 + 1 ln 7 2 3 (c) First, notice that the function ln( x ) is an increasing function of x , so that f ( x ) = 1 ln( x ) is a decreasing function of x . Since f ( x ) is decreasing, the right endpoint f ( x i ) is always less than the other values of f ( x ) in the subinterval [ x i - 1 , x i ], and the left endpoint f ( x i - 1 ) is always greater than the other values in the subinterval. Thus the

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