Note: Final Exam is at 10:45 on Tuesday, 5/3/11 (This is the Final Exam time reserved for our labs). From Practice Test I

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MA Practice Final Answers in Red 4/8/ and 4/9/ Name Note: Final Exam is at :45 on Tuesday, 5// (This is the Final Exam time reserved for our labs). From Practice Test I Consider the integral 5 x dx. Sketch a graph of this function and the area corresponding to this integral. Draw in five rectangles with equal widths and upper-right corners on the graph. We have been calling the sum of the areas of these rectangles the Right Sum with five rectangles. Use these five rectangles to compute an estimate for the given integral. (.6).6 + (.).6 + (.8).6 + (4.4).6 + (5.).6 = 6.44. The following is a program that will compute the Right Sum with rectangles. Fill in the missing step(s) between the do and the o f := x -> *x^; s := ; for n from by to do s := s + f (+n*.)*.; od; If we were to graph a speed/velocity function v(t) (measured in feet per seconds) with respect to t (measured in seconds), what units would we use to measure area in this graph? feet Draw a picture of the region that corresponds to x dx. e. You can use the area formula for a triangle to evaluate the integral in the previous problem. What is the base, and what is the height of the triangle? The base is, and the height is f() = 6.. Evaluate the following definite integrals using the Fundamental Theorem of Calculus. x dx. = x = 8. π/ cos(x) dx. = sin(x) π/ =. ex dx. = e x = e. x dx. = ln(x) = ln(). 4. Evaluate the following indefinite integrals. (Don t forget the +C s.) x + x + x + dx. = x4 4 + x + x + x + C. sin(x) dx. = cos(x) + C.

MA Practice Final Answers in Red e x + cos(x) dx. = e x + sin(x) + C. x dx. = x / dx = x/ / + C = x/ + C. 5. Use substitution to evaluate the following indefinite integrals. () x cos(x ) dx. Let u = x, then du = x dx. x cos(x ) dx = cos(u) du = sin(u) + C = sin(x ) + C. () x (x + 7) dx. Let u = x + 7, then du = x dx. x (x + 7) dx = u du = u 4 4 + C = (x + 7) 4 + C. () (x + ) x + x + dx. Let u = x + x +, then du = x +. (x + ) u x u / + x + dx = du = / + C = (x + x + ) / + C (4) xe x dx. Let u = x, then du = x dx. xe x dx = e u du = ex + C. 6. For each of the following, make the indicated u-substitution to rewrite the given definite integral in x as a definite integral in u. x(x + ) 4 dx, using u = x +. = 7 xex dx, using u = x. = 5 9 eu du. x x 4 dx, using u = x 4. = u du. u4 du. π/ cos(x)e sin(x) dx, using u = sin(x). = eu du.. From Practice Test II I said in class that all exponential functions can be expressed in terms of the exponential function e x. For example, the function f(x) = x could be written in the form f(x) = e rx for some constant r. Find r. We know that a = e ln(a), therefore, a x = e ln(a) x. In this case, r = ln().986 f := x -> x^; a := ; b := 4; N := 5; w := (b-a)/n; s := f(a)+f(b); for n from by to N- do s := evalf( s + *f(a+n*w) ); od;

MA Practice Final Answers in Red evalf(s*w/); In the program above, what definite integral are we approximating? 4 x dx. Which program is this? (I.e., is this a Right Sum, a Left Sum, et?) This is Trap(5). You can tell by the multipliers of on the middle terms, and the fact that we re dividing everything by. The following five numbers are the results of Right(5), Left(5), Trap(5), Mid(5), and Simp(5): 6.56494, 6.75495, 6.99494, 6.75, and 6.749949. Which one is which? Over this interval, f(x) = x is increasing, so Right(5) is an overestimate. We also know that Right and Left should be the worst approximations. The curve is concave up, so Trap(5) is an overestimate, and also Mid(5) and Trap(5) should be the third and fourth best estimates. The numbers are Right(5)=6.99494, Left(5)=6.56494, Trap(5)=6.75495, Mid(5)=6.749949, and Simp(5)=6.75. 5. Evaluate the following integrals. [[Note: I will give you a formula sheet covering the integrals we ve derive]] a) arctan(x) + C. b) arcsin(sin(x)) + C = x + C. +4x dx. cos(x) sin (x) dx. 6. Use integration by parts to do the following integrals. a) x ln(x) x + C. b) xex ex 4 + C. c) xe x e x + C. d) xsin(x) + cos(x) + C. 4 4 ln(x) dx. (Let u(x) = ln(x) and v (x) =.) xe x dx. xe x dx. x cos(x) dx. 8. Evaluate the following improper integrals. a) x d) x = 5 =. =. e) 5 5 x x dx. =. b) x =. c) e x = + =. x dx. e x dx. x / dx. e. x 4/5 dx. From Test III. Suppose you have an integral b f(x) dx, and you know that f(x) is concave down over the interval a [ a, b ]. Suppose also that MID() and TRAP() gave the two numbers.4475 and.4487, not necessarily in that order. Which of the two numbers is MID()? If the graph is concave down, then MID is an overestimate, so we re looking for the larger number,.4475.

MA Practice Final Answers in Red 4 You know absolutely for sure that the actual value of the integral, rounded to one decimal place, is.4. Right? Based on the information you have, what is the maximum number of decimal places that you can round the actual value of the integral to, and be absolutely sure that you ve got the right number? Three. If you round the two numbers to three decimal places, you get the same thing, but rounding to four gives you different numbers. What s that number?.4. You have some function f, and you re trying to approximate the integral 4 f(x) dx using the Monte Carlo metho You know that < f(x) < 5, so you test 5 randomly generated points in the rectangle x 4 and y 5. If 44 of these points lie below the curve, what is the corresponding approximation of 4 f(x) dx? The proportion of dart hits below the curve approximates the proportion of area below the curve. Therefore, our approximation is 44 5 =.6. 5. Use L Hôpital s rule on the following limits. x lim + 6x x. = lim 5x +7x x+7 = lim 6 = 5. lim x x e x. = lim x e x = lim e x =. 4. Evaluate the following improper integral. 4x 4 x dx. /4 = 4. 5. Evaluate the integral sin (θ) dθ. cos(θ) dθ = θ sin(θ) 4 + C. 6. Evaluate the integral x dx using the substitution x = sin(θ). sin (θ) cos(θ) dθ = 9π 4. π/ 8. Find the two basic solutions to the given differential equation. y 4y 5y =. Solve r 4r 5 = to get r =, 5 and the two solutions y = e x, e 5x. y + y =. Solve r + r = to get r =, and the two solutions y =, e x. 9. Solve this separable differential equation, xy + (x + )y =. solution is y = e x ln(x) + C = C xe. x y dy = x dx. The explicit. Find the area between y = sin(x) and y = x π shown below. π/ sin(x) x π dx = π 4 +.

MA Practice Final Answers in Red 5. Suppose you had a square-based pyramid of height 5, and the base has sides of length. Use an integral ( to find the volume (i.e., using the volume formula for a pyramid will count for nothing). 5 x ) dx = 5. 5. We have a conical tank filled with water. The tank is 6 feet deep and feet across at the top. Compute the work necessary to empty the tank. (The density of water is 6.4 pounds per cubic foot.) 6 (6 x)6.4π ( 5 6 x ) dx = 4,7.7.

MA Practice Final Answers in Red 6 After Test III. For the differential equation dy dx = x y with the initial condition y() =, use Euler s method to find y(.6) using steps of dx =.. At the point (, ), the slope is dy dx = () =, so as we move from x = to x =., y changes dy =. =. The next point is (., ). At the point (., ), the slope is dy dx = (.) =.8, so as we move from x =. to x =.4, y changes dy =.8. =.6. The next point is (.4,.6). At the point (.4,.6), the slope is dy dx = (.4).6 =.565, so as we move from x =.4 to x =.6, y changes dy =.565. =.645. The next point is (.,.85). The approximation of y(.6) is.85.. Find the best-fit cubic p(x) = ax + bx + cx + d to the function f(x) = e x at x =. We know that p() = d, p () = c, p () = b, p () = 6 We also know that f() =, f () =, f () = 4, f () = 8. This means that (5) p(x) = 4 x + x + x +.. Find the sum of the following series. ( i= ) i. =. ( i= ) i. = 4. 4. Determine whether the following series converge or diverge, and tell how you know (e.g., converges by the p-test, because p = >. ) i= i. Converges by p-test, because p = >. i= i. Converges because this is a geometric series with r =. i= i +. This is smaller than the p-series i, which converges, because p = >. Our series converges, therefore, by the comparison test. e. f. i= i i!. By the ratio test, we have a ratio i+ <, so this series converges. i= i= ( ) i i+. Converges by the alternating series test, since the terms go to zero. ( ) i. The terms go to, so this series diverges by the divergence test. i 5. Find the interval and radius of convergence for each of the given power series. i= (x) i i. By the ratio test, the ratio is x i i+ x <, so the series converges for < x <. For x =, we have an alternating harmonic series, which converges. For x =, we have the harmonic series, which diverges. Therefore, the interval of convergence is (6) x <, and the radius of convergence is. i= xi x i. By the ratio test, the ratio is x <, so the series converges for < x <. The i (i+) series also converges at the endpoints (alternating series test and p-series, with p = > ), so the interval of convergence is (7) x,

MA Practice Final Answers in Red 7 and the radius of convergence is. (x ) i i= i. By the ratio test, the ratio is (x ) i i+ x <, so the series converges for < x <. We have an alternating harmonic series at x =, and a harmonic series at x =, so the interval of convergence is (8) x <, with a radius of convergence of.