To calculate the x=moment, we multiply the integrand above by x to get = )

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1 . ( points) The region shown below is the area between the curves y = 3x and y = x. Find the center of mass of this region We must calculate the area, x-moment, and y-moment to find the center of mass. The region is bounded on the left by x = and on the right by x = 3. The upper curve is y = 3x and the lower curve is y = x. Thus, the integral to calculate the area is A = ] 3 ( 3x x = 3x x3 7 = 3 7 ) ( ) = 9 3 To calculate the x=moment, we multiply the integrand above by x to get M x = ] 3 ( 3x x 3 = x 3 x = 7 8 ) ( ) = 7 To calculate the y-moment, however, we need to use one half the difference of the squares of the upper and lower functions: M y = (3x) ( ) x = 9x x = Thus, the center of mass of the above region is ( Mx, My ) = ( 3, ) 8 A A. ] 3 [3x 3 x = ( 8 3 ) ( ) = 8. ( points) Find the general solution to the differential equation dy xy = ex x. This is a linear differential equation, with coefficient x on the y-term. Thus, we calculate the integrating factor: ρ = e R x = e x Multiplying the whole equation by this integrating factor, we see (or hope, at least) that the left side folds into a single derivative, which we can then integrate: dy x e xe x y = e x x e x d ( ) e x y = e x e x y = e x e x y = e x + C y = e x x + Ce x Page of October 3, 8

2 3. ( points) Evaluate the following integrals, or if they cannot be evaluated, demonstrate why not. (a) (7 points). x The function has a discontinuity at x =, so the above integral is improper and must x be rephrased as a sum of limites of definite integrals: b + lim x b x a + a x ln x ]b b + lim ln x ] a + a ln b ln + lim ln ln a b a + = ln ln + lim ln b lim ln a b a + Since zero has no natural logarithm, or and there is not even a limit to the natural logarithm as its argument approaches zero, neither of the limits shown above exist, so the integral is divergent. (b) (8 points) 3 x. Using limits to rephrase this improper integral: 3 x b (x ) /3 (x ) /3 /3 (b ) /3 /3 ( )/3 /3 However, as b grows without bound, so does (b ) /3, so this integral is divergent. for x <. ( points) Consider the function f(x) = k with k a constant. for x x/ (a) (6 points) Find a value of k such that f(x) is a probability distribution function. A cursory inspection reveals that this function is non-negative throughout if k is positive: is non-negative everywhere, and is non-negative in its entire domain. The critical x / property to demonstrate that this function is a probability distribution function is simply that f(x) =. We can simplify this somewhat by ignoring the region on which Page of October 3, 8

3 f(x) is zero, so that f(x) = f(x). We evaluate this as such: f(x) = b = + k k x / k 3x 3/ k x / k 3b 3/ + k 3 3/ so for k to equal, we would choose k =. (b) (6 points) For a random variable X described by the above probability distribution function, find the average value of X. The expected value (or average value) of a probability distribution function f(x) is xf(x). We may ignore locations where the integrand is zero, so this can be simplified to xf(x): xf(x) = b x x / x / x 3/ + b/ = + = (c) (3 points) For a random variable X described by the above probability distribution function, find P(X 9). This probability will be simply 9 f(x). It is techincally 9 f(x), but we can ignore the region over which the function is zero. 9 f(x) = 9 8 x/ = x 3/ ] 9 / = = 3/ 3/ 7 + = 9 7. ( points) Consider the curve y = e x + between the points (, ) and (, + e ). (a) ( points) Construct, but do not evaluate, an integral representing the length of this curve. The arclength expression + ( ) dy evaluates in this case to + (e x ), so the arclength is + e x Page 3 of October 3, 8

4 (b) (3 points) Construct, but do not evaluate, an integral representing the surface area of the surface produced by rotating this curve around the vertical line x = 3. Such a revolution would involve arcs given, as shown in part (a), to be of differential length + e x, being spun around circles of radius x+3, since the horizontal distance between the line x = 3 and the point (x,e x + ) is x + 3; thus the differential area traced out is π(x + 3) + e x, so the integral to compute the total surface area is π(x + 3) + e x (c) (3 points) Construct, but do not evaluate, an integral representing the surface area of the surface produced by rotating this curve around the x-axis. Such a revolution would involve arcs given, as shown in part (a), to be of differential length + e x, being spun around circles of radius e x +, since the vertical distance between the line y = and the point (x,e x +) is simply e x +; thus the differential area traced out is π(e x + ) + e x, so the integral to compute the total surface area is π(e x + ) + e x 6. ( points) Perform the approximations shown below. (a) ( points) Using Simpson s rule with n = 6, approximate. You need not arithmetically simplify your result. x Since the interval has length 3 and n = 6, the choice of x is 3 =. Thus, we will need 6 to sample the integrand at the seven points x =, 3,,, 3, 7,. Assembling these into Simpson s rule, we would find that the approximation is: ( / 3 + 3/ + + / / + ) If this were simplified, we would have 397, which is not a particularly bad rational estimate for the actual integral, which is ln. (b) ( points) Consider the differential equation dy = xy subject to the initial condition that when x =, y =. Using Euler s method with a step size of., approximate the value of y when x =. dy We have a differential equation whose slope (i.e. ) at each point is described by the function m(x,y) = xy. We will be using Euler s method on this with x =. and initial point of (x,y ) = (, ). From this, we will calculate new positions x and y. x = x + x = +. =. y = y + xm(x,y ) = +. ( ) = so the second point in our estimation of this curve is (., ). We repeat Euler s method at this new point to find x and y : x = x + x =. +. = y = y + xm(x,y ) = +. (.)( ) = 6 Page of October 3, 8

5 so the third point in our estimation of this curve is (, 6); thus, when x =, we estimate y to be 6. Note that this is a separable differential equation: were it to be solved, one would get the result y = e x, so the correct value of y when x = is in fact e 6, which is far less than ( points) Answer the following questions about the differential equation dy = ex y. (a) ( points) Demonstrate without explicitly solving the differential equation that y = e.x is a solution. Evaluating the left and right sides of the differential equation, and substituting in this solution, we get the two expressions: d e.x = (.)e.x e x e.x = e.x which are identical, given that (.) = =. (b) ( points) Find the general solution of the differential equation. Using separable methods, we rearrange this integral into y dy = ex ydy = e x y = e x + C y = e x + C y = e x + C Note that the specific solution given in part (a) is this general solution with C =. (c) ( points) Using your general solution, find a solution to the differential equation subject to the initial condition that y = e when x =. Plugging y = e and x = into the solution found above: e = e + C, leading to 6e = e + C, so C = e, leading to the specific solution y = e t + e. Page of October 3, 8