Exam 14: Further Integration Techniques and Applications of the Integral

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Evaluate the integral. 01x2(x+2)2 dx\int _ { 0 } ^ { 1 } x ^ { 2 } ( x + 2 ) ^ { 2 } \mathrm {~d} x

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Find the total value of the given income stream and also find its present value (at the beginning of the given interval) using the given interest rate. R(t)=50,000R ( t ) = 50,000 , 0t150 \leq t \leq 15 , at 9%

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Evaluate the integral. (5z2ez5zez2)dz\int \left( 5 z ^ { 2 } e ^ { z } - 5 z e ^ { z ^ { 2 } } \right) \mathrm { d } z

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Nancy's Chocolates estimates that the elasticity of demand for its dark chocolate truffles is E=0.06p2E = 0.06 p - 2 Where p is the price per pound. Nancy's sells 19 pounds of truffles per week when the price is $23 per pound. Find the formula expressing the demand q as a function of p. Recall that the elasticity of demand is given by E=dq dp×pqE = - \frac { \mathrm { d } q } { \mathrm {~d} p } \times \frac { p } { q } . NOTE: Round the value of the constant in your answer to four decimal places.

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Evaluate the integral. x2ln(2x)dx\int x ^ { 2 } \ln ( 2 x ) \mathrm { d } x

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Calculate the 7-unit moving average of the function. f(x)=e0.5xf ( x ) = e ^ { 0.5 x }

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Calculate the 12-unit moving average of the given function. ​ f(x)=x39xf ( x ) = x ^ { 3 } - 9 x

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Find the area of the region enclosed by y=2x3y = 2 x ^ { 3 } and y=2x4y = 2 x ^ { 4 } .

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For the given differential equation, find the particular solution y=9ln2y = - 9 \ln 2 when x=ln2x = - \ln 2 . dy dx=9ex\frac { \mathrm { d } y } { \mathrm {~d} x } = 9 - e ^ { - x }

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For the differential equation, find the particular solution if y=12y = \frac { 1 } { 2 } when x=15x = \frac { 1 } { 5 } . dy dx=y2x2\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y ^ { 2 } } { x ^ { 2 } }

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The rate of U.S. per capita sales of bottled water for the period 1993-2003 could be approximated by Q(t)=0.05t2+0.4t+9Q ( t ) = 0.05 t ^ { 2 } + 0.4 t + 9 where t is the time in years since 1990. Compute the average annual sales of bottled water over the period 1996-2003, to the nearest gallon per year.

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A fast-food outlet finds that the demand equation for its new side dish, "Sweetdough Tidbit", is given by p=6,750(q+5)2p = \frac { 6,750 } { ( q + 5 ) ^ { 2 } } where p is the price (in cents) per serving and q is the number of servings that can be sold per hour at this price. At the same time, the franchise is prepared to sell q=0.5p5q = 0.5 p - 5 servings per hour at a price of p cents. Find the consumers' surplus at the equilibrium price level.

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Decide whether the integral converges. If the integral converges, compute its value. 159x13 dx\int _ { 1 } ^ { 5 } \frac { 9 } { x ^ { \frac { 1 } { 3 } } } \mathrm {~d} x

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Find the area of the region enclosed by y=6lnxy = 6 \ln x , y=7lnxy = 7 - \ln x , and the line x=6x = 6 .

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Find the area of the region between y=17xy = \frac { 1 } { 7 } x and y=17x2y = \frac { 1 } { 7 } x ^ { 2 } for x in [1,1][ - 1,1 ] .

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Find the total value of the given income stream and also find its present value (at the beginning of the given interval) using the given interest rate. R(t)=90,000e0.03tR ( t ) = 90,000 e ^ { 0.03 t } , 0t50 \leq t \leq 5 , at 6%

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The annual net sales (revenue) earned by Nintendo Co., Ltd., in the fiscal years ending March 31, 1995 to March 31, 2003, can be approximated by R(t)=4t2+10t+530R ( t ) = - 4 t ^ { 2 } + 10 t + 530 billion yen per year (6t3)( - 6 \leq t \leq 3 ) where t is time in years ( t=0t = 0 represents March 31, 2000). Suppose that, from April 1994 on, Nintendo invested its revenue in an investment yielding 4% compounded continuously. What, to the nearest 100 billion, would the total value of Nintendo's revenue have been by the end of March 2003

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Find the area of the region enclosed by y=exy = e ^ { x } , y=2y = 2 , and the y-axis.

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Evaluate the integral. (x2+1)e8x+2 dx\int \left( x ^ { 2 } + 1 \right) e ^ { 8 x + 2 } \mathrm {~d} x

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Calculate the 11-unit moving average of the function. f(x)=xf ( x ) = \sqrt { x }

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