x+1 e 2t dt. h(x) := Find the equation of the tangent line to y = h(x) at x = 0.
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1 Math Sample final problems Here are some problems that appeared on past Math exams. Note that you will be given a table of Z-scores for the standard normal distribution on the test. Don t forget to have a serious look at HW9. () (a) State both parts of the Fundamental Theorem of Calculus as accurately as possible. Section 5.4 (b) Consider the function defined by h(x) : x+ e t dt. Find the equation of the tangent line to y h(x) at x. To compute the tangent line we need h() and h (). For the first h() e t dt et ( e ) x. If we let F (x) e t dt, then F (x) e x, h(x) F (x+), and by the Chain Rule h (x) d F (x+) F (x+ dt ) d (x + ) dt e(x+), so h () e. The tangent line is y e x + (e ). (c) On what subinterval(s) of [, ), if any, is the graph of y h(x) increasing? Explain. its domain. h (x) e (x+) > for all x, and so h is increasing everywhere on () Consider the function f(x) equal to 4x( x ) if x and equal to otherwise. (a) Show that f(x) is a probability density function. In order for f to be a probability distribution function it needs to satisfy two conditions: ) f(x) >, and ) f(x). ) can be checked by looking at the graph or by noting that on [,], x is positive. To check ) f(x) 4x( x ) 4x 4x 3 x x 4 ( ) ( ) (b) Compute P ( X /).
2 Math Final March P ( X /) / / f(x) 4x 4x 3 (x x 4 ) (c) Compute the mean of f(x). / (/ /6) 7/6 The mean µ of X is xf(x), so µ xf(x) x4x( x ) 4x 4x 4 4/3x 3 4/5x 5 (4/3 4/5) 8/5 (3) (a) The probability density function for a random variable X that is normally distributed with mean µ and standard deviation σ is given by f(x) σ (x µ) π e σ. Show using the change of variable Z (X µ)/σ that Z is normally distributed with mean and standard deviation. To determine whether Z is normal and what its mean and standard deviation are, we need to compute a probability density function for Z. In other words, we need an integral to compute P (A Z B). The substitution, though, says that P (A Z B) P (A (X µ)/σ B). Since the conditions A (X µ)/σ B give Aσ + µ X Bσ + µ, this is the same Bσ+µ as P (Aσ + µ X Bσ + µ) or Aσ+µ σ π e (x µ) σ. Applying the u- subsitution z (x µ)/σ, dz /σ gives z A when x Aσ + µ and z B when x Bσ + µ and so Bσ+µ Aσ+µ σ π e (x µ) σ Bσ+µ Aσ+µ B A π e ( x µ σ π e (z) dz ) σ
3 Math Final March So, P (A Z B) B A π e z dz and π e z is the probability density function for Z. Comparing this with the general formula, we see that Z is a normally distributed random variable with mean µ and standard deviation σ. (b) A radar unit is used to measure speeds of cars on a motorway. The speeds are normally distributed with a mean of 9 km/hr and a standard deviation of km/hr. What is the probability that a car picked at random is travelling at more than km/hr? We need to compute P (X ). The Z-score for is ( 9)/ so we need to know P (X µ + σ). Using the transformation from part i) (Z (X 9)/), P (X ) P (Z ) P (Z ). Checking the normal distribution table, shows P (Z ).843 and so P (X ) P (Z ) Thus the probability that someone is traveling over kmh is 5.87%. (4) A football is thrown upward from a height of 4 feet with a velocity of feet per second. Assume the acceleration is a constant 3 ft/sec acting down. Determine the height of the football at any time and the time at which the ball hits the ground. v(t) a(t) 3 a(t) 3t + D p(t) D v() v(t) 3t + v(t) 6t + t + C C p() 4 p(t) 6t + t + 4 The height of the ball is given by p(t) 6t + t + 4. The ball hits the ground when p(t) 6t + t + 4 4t + 5t + Using the quadratic equation gives t 5± 5 4( 4)() 8. The positive root is (5 + 4)/8. The ball hits the ground after (5 + 4)/8 seconds. 3
4 Math Final March (5) Set up but DO NOT evaluate an integral computing each of the volumes below. Use the method of your choice. (a) the volume of the solid obtained by rotating the region bounded by y x + and y 9 x about y. First we need to check that the two curves intersect at two points - (,5) and (-,5). We ll use the washer method. The inner and outer radii point vertically and we write their heights as a function of x. Volume b a πr out πr in π((9 x ) ( )) π((x +) ( )) (b) the volume of the solid obtained by rotating the region bounded by the x-axis, x y, x + y about the y-axis. π( x ) π(x +) First we need to check that the two curves intersect at the point (,). We ll use the washer method, integrating in the y direction. The inner and outer radii point horizontally and we write their lengths as a function of y. 4
5 Math Final March Volume b πr out πr in dy a π( y) π(y) dy (6) Derive the formula for the volume of a sphere of radius r. This is example of 6. (7) Sketch the region bounded by the curves x, y, 6x y 3 and x + y 4, and then compute its area. The line 6x 3 crosses the x-axis at (/,), and intersects the graph of 4 x at (,3). So, we need to break the area integral into two regions. On [, /], we look at the region between the x-axis and 4 x. On [/, ], we look at the area between 4 x and 6x 3. The total area is: Area / 4 x + (4x 3 x3 ) / 35/ 5 / (4 x ) (6x 3) + (7x 3x 3 x3 ) /
6 Math Final March (8) Integrate. Remember to interpret improper integrals as necessary. xe x xe x x 4 x 9 x 3x x x + x + x x 5x + x 3 + x sin 5 (x) (a) so xe x lim t t xe x xe x xex 4 ex t lim t xe x lim ( t xex 4 ex ) lim t ( tet 4 et ) ( ()e() 4 e() ) ( 3 4 e ) 3 4 e t (b) let u x, du x, then xe x e u du e x + C (c) x 4 x + 4 x 4 lim t t x 4 x 4 lim t 3 x 3 t lim t 3 t 3 4 This diverges. 6
7 Math Final March (d) + x + x + x + + x tan x x + x For, x + x, let u + x, du x x + x u du ln u ln + x (e) (f) So + x + x tan x + ln + x + C x 9 x 3x x 3x (x 5)(x + ) x 9 (x 5)(x + ) A x 5 + B x + x 9 A(x + ) + B(x 5) Substituting x gives B 3, and x 5, gives A so x 9 x 3x x ln x ln x + + C x + So A, B 5. x 5x + x 3 + x x 5x + x 3 + x A x + B x + x 5x + A(x + ) + B(x) (g) x 5x + x 3 + x x + 5 x + ln x 5 tan x + C x x x sin θ, cos θ dθ 7
8 Math Final March x sin θ x cos θ sin θ dθ cos θ dθ cos θ dθ θ sin θ + C 4 θ sin θ cos θ + C 4 sin x x x + C (h) sin 5 (x) ( cos x) sin x Let u cos x, du sin x, then ( u ) du u + u 4 du u + 3 u3 5 u5 + C cos x + 3 cos3 x 5 cos5 x + C 8
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