Exam1_S2010 Solutions

# B z cos 8 x sin x dx use u substitution choose u cos

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b. Z cos 8 ( x ) sin( x ) dx Use u -substitution. Choose u = cos( x ). Then du = - sin( x ) dx . Changing variables, we get Z cos 8 ( x ) sin( x ) dx = - Z u 8 du = - 1 9 u 9 + C = - 1 9 cos 9 ( x ) + C c. Z ln( x ) x 2 dx Use integration by parts. Choose u = ln( x ) and dv = 1 /x 2 = x - 2 dx . Then du = 1 x dx and v = - 1 x . Integration by parts gives: Z ln( x ) x 2 dx = - ln( x ) x - Z - 1 x · 1 x dx = - ln( x ) x + Z x - 2 dx = - ln( x ) x - x - 1 + C 2

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5. Find the average value of f ( x ) = sec 2 ( x ) tan( x ) on the interval [ π/ 4 , π/ 3] . Recall the average value of a function f ( x ) on [ a, b ] is given by 1 b - a Z b a f ( x ) dx . Thus we must first compute the indefinite integral Z sec 2 ( x ) tan( x ) dx . We use u -substitution with u = tan( x ). Then du = sec 2 ( x ) dx . Changing variables, we have: Z sec 2 ( x ) tan( x ) dx = Z u du = 1 2 u 2 + C = 1 2 tan 2 ( x ) + C. Taking C = 0, we evaluate the definite integral: Z π/ 3 π/ 4 sec 2 ( x ) tan( x ) dx = 1 2 tan 2 ( x ) π/ 3 π/ 4 = 1 2 tan 2 ( π/ 3) - 1 2 tan 2 ( π/ 4) . Recall that tan( x ) = sin( x ) cos( x ) . Thus tan( π/ 3) = 3 2 · 2 1 = 3 and tan( π/ 4) = 2 2 · 2 2 = 1. Thus: Z π/ 3 π/ 4 sec 2 ( x ) tan( x ) dx = 1 2 ( 3) 2 - 1 2 (1) 2 = 3 2 - 1 2 = 1 . Thus the average value of f is 1 π/ 3 - π/ 4 . 6. Find the present value of an annuity if the monthly payments are \$680 per month for 18 years, and the account earns interest at a rate of 7% per year. Recall the present value of an annuity is given by PV = mP r ( 1 - e - rT ) . Filling in our information, we get PV = (12)(680) 0 . 07 1 - e - (0 . 07)(18) . 7. Find the relative extrema of the function f ( x ) = sin( x ) cos( x ) on the interval (0 , π ). To find the extrema of a continuous function, we must find its critical points. These are the solutions to the equation f 0 ( x ) = 0, so first we use the product rule to find: d dx sin( x ) cos( x ) = cos 2 ( x ) -
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