Math 181, Exam 2, Study Guide 2 Problem 1 Solution. 1 + dx. 1 + (cos x)2 dx. 1 + cos2 xdx. = π ( 1 + cos π 2

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1 Math 8, Exam, Study Guide Problem Solution. Use the trapezoid rule with n to estimate the arc-length of the curve y sin x between x and x π. Solution: The arclength is: L b a π π + ( ) dy + (cos x) + cos x We now use the trapezoid rule with n to estimate the value of the integral. The formula we will use is: T x ( π ] f() + f + f(π) ) where f(x) + cos x and the value of x is: The value of T is then: x b a n π π T x ( π ] f() + f + f(π) ) π + cos + + cos π + ] + cos π π ] π ( + ) 4

2 Math 8, Exam, Study Guide Problem Solution. (a) Let R be the region between y +x and the x-axis with x. Does R have finite area? If so, what is the area? (b) Let S be the solid obtained by revolving R around the y-axis. volume? If so, what is the volume? Does S have finite Solution: (a) The area of R is given by the improper integral: Area + x + We evaluate the integral by turning it into a limit calculation. + x + R lim R + The integral has a simple antiderivative so its value is: R x + ] R x + arctan x arctan R arctan arctan R We now take the limit of the above function as R +. + x + lim R + R x + lim R + arctan R π Thus, the area is finite and its value is π. (b) The volume of S is obtained by using the Shell Method. The formula is V πx x +

3 To compute the integral we first turn it into a limit calculation. The value of the integral is The volume is then b That is, the volume is not finite. b x V π lim b x + x x + ln x + ] b ln(b + ) V lim b, ln(b + )

4 . Evaluate the following integrals: (a) (b) (c) (d) π π Solution: π π sin 4 x x + 5x + x + 4x + x e x Math 8, Exam, Study Guide Problem Solution (a) We solve this integral using a reduction formula. sin n x n sinn x cos x + n n sin n x Letting n 4 we get: sin 4 x 4 sin x cosx + sin x 4 4 sin x cosx + cos(x)] 4 4 sin x cosx + cos(x)] 8 4 sin x cosx + 8 x 6 sin(x) To evaluate sin x we used the double angle identity cos(x) sin x. We now solve the definite integral. 4 sin x cosx + 8 x 6 sin(x) ] π π sin 4 x 4 sin π cosπ + 8 π + 8 ] π 8 ] π ] 6 sin(π) 4 sin ( π) cos( π) + 8 ( π) ] 6 sin( π) 4 π

5 (b) We will evaluate the integral using Partial Fraction Decomposition. First, we factor the denominator and then decompose the rational function into a sum of simpler rational functions. x + 5x + (x + )(x + ) A x + + B x + Next, we multiply the above equation by (x + )(x + ) to get: A(x + ) + B(x + ) Then we plug in two different values for x to create a system of two equations in two unknowns (A, B). We select x and x for simplicity. x ( : A ) ( ( + + B ) ) + A x : A( + ) + B(( ) + ) B Finally, we plug these values for A and B back into the decomposition and integrate. ( A x + 5x + x + + B ) x + ( ) x + + x + ln x + ln x + We now solve the definite integral. x + 5x + ] ln x + ] ln x + ln () + ln + ln ln ln + ln ln () + ] ln + ln (c) We begin by completing the square in the denominator. x + 4x + (x + ) +

6 We then evaluate the integral using the u-substitution method. Let u (x + ). Then du du and we get: x + 4x + We now solve the definite integral. x + 4x + arctan (x + ) + (x + )] + du u + du u + arctan u + C ] arctan (x + ) (x + ) ] ] ] arctan ( + ) arctan ( arctan ) ( )] arctan ( + ) ] (d) We evaluate the integral using Integration by Parts. Let u x and v e x. Then u x and v e x. Using the Integration by Parts formula: uv uv u v we get: x e x x e x x ( e x) x e x + xe x A second Integration by Parts must be performed. Let u x and v e x. Then u and v e x. Using the Integration by Parts formula again we get: ( e x e x x e x + xe x x ) ] x e x xe x + e x x e x xe x e x

7 We now solve the definite integral. We recognize that it is an improper integral so we turn it into a limit and evaluate. x e x lim b b x e x lim x e x xe x e x] b b lim b e b be b e b + ] b lim b b e b b e b e b + + When computing the limits above, we used the fact that: lim x by repeated application of L Hopital s Rule. x n e x ] 4

8 Math 8, Exam, Study Guide Problem 4 Solution 4. Use a Taylor polynomial for y e x to calculate e to two decimal places. Explain (using the remainder formula) why you have used sufficiently many terms. Solution: We will find the nth degree Maclaurin polynomial of f(x) e x so that the error T n () f() T n () e is less than. That is, we must find a value of n that ensures that the Error Bound satisfies the inequality: x a n+ Error T n () e K (n + )! < where x, a, and K satisfies the inequality f (n+) (u) K for all u, ]. Since f (n+) (u) e u < for all u, ] we choose K. We now want to satisfy the inequality: n+ Error T n () e < (n + )! Error T n () e (n + )! < We will find an appropriate value of n by a trial and error process. n 4 5 (n + )!!! 6 4! 8 5! 4 6! 4 < Therefore, we choose n 5. The Maclaurin polynomial T 5 (x) for f(x) e x is: Evaluating at x we get: T 5 (x) + x +! x +! x + 4! x4 + 5! x5 T 5 () + +! +! + 4! + 5! T 5 ()

9 Math 8, Exam, Study Guide Problem 5 Solution 5. Let S be the surface obtained by revolving the curve y sin x between x and x π around the x-axis. What is the surface area of S? Solution: The surface area formula is: Surface Area π b a f(x) + f (x) Using a, b π, f(x) sin x, and f (x) cosx we get: π Surface Area π sin x + (cosx) π π sin x + cos x To evaluate the integral we use the u-substitution u cosx. Then du sin x, the lower limit of integration changes from to, and the upper limit changes from π to. Making these substitutions we get: Surface Area π π 4π + u du + u du + u du where, in the last step, we used the fact that + u is symmetric with respect to the y-axis. To evaluate this integral we ll use the trigonometric substitution u tan θ and du sec dθ. We get: + u du + tan θ sec θ dθ sec θ sec θ dθ sec θ dθ tanθ sec θ + ln sec θ + tanθ where the integral of sec θ is determined via a reduction formula. Using the fact that u tanθ we find that sec θ + u using a Pythagorean identity. Therefore, + u du u + u + ln + u + u

10 The surface area is then: Surface Area 4π + u du 4π u + u + ln ] + u + u 4π () + + ln ] + + ] ] 4π + ln 4π + ln π + π ln 4π () + + ln ] + +

11 Math 8, Exam, Study Guide Problem 6 Solution 6. (a) Estimate ln using the degree two Taylor polynomial for y ln x around x. (b) Estimate ln using the Midpoint rule with n for the integral / (c) Calculate the error bounds for the two estimates. Does this tell you which is closer to the exact answer? Solution: (a) The degree two Taylor polynomial for f(x) ln x around x has the formula: T (x) f() + f ()(x ) + f () (x )! The derivatives of f(x) evaluated at x are: x. The Taylor polynomial T (x) is then: k f (k) (x) f (k) () ln x ln x x T (x) f() + f ()(x ) + f () (x )! T (x) + (x ) (x )! T (x) (x ) (x ) We will now estimate ln using T ( ). ln ( ) T ( ) ( ) 8

12 (b) The value of x in the Midpoint rule is: The Midpoint estimate M is: x b a n 4 M x f ( ) 9 8 ] + f ( )] ] (c) The error bound for part (a) is given by the formula: x a n+ Error K (n + )! where x, a, n, and K satisfies the inequality f (u) K for all u, ]. One can show that f (x) x. We conclude that f (u) u < for all u, ]. So we choose K and the error bound is: Error! 4 The error bound for part (b) is given by the formula: Error(M n ) K(b a) 4n where a, b, n, and K satisfies the inequality ( x ) K for all x, ]. We conclude that ( x ) for all x, ]. So we choose K and the error x bound is: Error(M ) ( ) 4() 84 We cannot tell which of 4 and is closer to the exact answer. All we know is that 8 99 both errors are smaller than. 4

13 Math 8, Exam, Study Guide Problem 7 Solution 7. Does the improper integral + Solution: We begin by rewriting the integral as follows: + converge or diverge? Justify your answer. + x + x + + x + + x The first integral on the right hand side is a proper integral so we know that it converges. We will use the Comparison Test to show that the second integral converges. Let g(x). +x We must choose a function f(x) that satisfies: () + f(x) converges and () g(x) f(x) for x We choose f(x) x. This function satisfies the inequality: g(x) f(x) + x x for x because the denominator of g(x) is greater than the denominator of f(x) for these values of x. Furthermore, the integral + f(x) + converges because it is a x p-integral with p >. Therefore, the integral + g(x) + converges by +x the Comparison Test and the integral + converges. +x

14 Math 8, Exam, Study Guide Problem 8 Solution 8. What is the arc-length of the segment of the parabola y 4 x above the x-axis? Solution: The arclength is: L b a + f (x) + ( x) + 4x + 4x We solve the integral using the trigonometric substitution x tanθ, sec θ dθ. The indefinite integral is then: + 4x ( ) ( ) + 4 tan θ sec θ dθ + tan θ sec θ dθ sec θ sec θ dθ sec θ dθ 4 tanθ sec θ + ln sec θ + tanθ 4 Using the fact that x tanθ we find that tanθ x and sec θ + 4x either using a triangle or a Pythagorean identity. The integral in terms of x is then: + 4x 4 tanθ sec θ + 4 ln sec θ + tan θ x + 4x + 4 ln + 4x + x

15 The arclength is then: L + 4x ] x + 4x + 4 ln + 4x + x x + 4x + ln ] + 4x + x + 4() + ln ] + 4() + () + 4() + ln ] + 4() + () 7 + ln ( )

16 Math 8, Exam, Study Guide Problem 9 Solution 9. Find a formula for the general Taylor polynomial T n (x) for the following functions around the specified points: (a) e x around x (b) x around x Solution: (a) We ll use a shortcut to find T n (x). We ll start with the general Maclaurin polynomial for e x which is: n T n (x) + x + x! + x! + + xn n! x k k! and replace x with x to get: k T n (x) x + x4! x6! + + ( x ) n n! n ( x ) k k k! (b) The function f(x) and its derivatives evaluated at a are: The Taylor polynomial of degree n is: k f (k) (x) f (k) () x / x / 4 x / 8 x 5/ x 7/ 5 4 T n (x) + (x )! (x ) +! (x ) ! + n (x ) + ( ) k... (k ) (x ) k k k! k