PRELIM 2 REVIEW QUESTIONS Math 1910 Section 205/209

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1 PRELIM 2 REVIEW QUESTIONS Math 9 Section 25/29 () Calculate the following integrals. (a) (b) x 2 dx SOLUTION: This is just the area under a semicircle of radius, so π/2. sin 2 (x) cos (x) dx SOLUTION: sin 2 (x) cos (x) dx = ( cos 2 (x)) cos (x) dx = cos (x) dx cos (x) dx Now use the reduction formula. cos (x) dx = sin(x) cos (x) + cos 2 (x) dx = sin(x) cos (x) + ( + cos(2x)) dx 2 = sin(x) cos (x) + ( 2 x + ) 2 sin(2x) + C cos (x) dx = sin(x) cos5 (x) = sin(x) cos5 (x) + 5 cos (x) + 5 ( sin(x) cos (x) + ( 2 x + ) ) 2 sin(2x) + C Therefore, the answer is sin(x) cos (x) + ( 2 x + ) ( sin(x) cos 2 sin(2x) 5 (x) + 5 ( sin(x) cos (x) + ( 2 x + ) ) 2 sin(2x) + C (c) sin 5 (x) cos (x) dx SOLUTION: This one looks like the reduction formula, but it s just substitution! sin 5 (x) cos (x) dx = sin(x)( cos 2 (x)) 2 cos (x) dx ( ) = sin(x) cos (x) 2 cos 2 (x) + cos (x) dx = sin(x) cos (x) 2 sin(x) cos (x) + sin(x) cos 8 (x) dx

2 Set u = cos(x), du = sin(x) dx. Therefore, we get sin 5 (x) cos (x) dx = u 2u + u 8 du = u u7 7 u9 9 + C = cos5 (x) cos7 (x) 7 cos9 (x) 9 + C (d) tan (x) sec (x) dx SOLUTION: tan (x) sec (x) dx = sec 2 (x) tan (x)(tan 2 (x) + ) dx Let u = tan(x), du = sec 2 (x) dx. tan (x) sec (x) dx = u (u 2 + ) du (e) cot 5 (x) csc 5 (x) dx SOLUTION: cot 5 (x) csc 5 (x) dx = = u9 9 + u7 7 + C = tan(x)9 9 + tan(x)7 7 + C cot(x) csc(x)(csc 2 (x) ) 2 csc (x) dx Let u = csc(x), du = cot(x) csc(x) dx. cot 5 (x) csc 5 (x) dx = = = = csc9 (x) 9 (u 2 ) 2 u du (u 2u 2 + )u du u 8 2u + u du + 2 csc7 (x) 7 csc5 (x) 5 + C (f) x x 2 dx

3 SOLUTION: Let x = 2 sin θ, dx = 2 cos θ dθ. x dx = x 2 2 sin θ (2 cos θ dθ) sin 2 θ = 2 sin θ dθ = 2 cos θ + C = 2 x 2 + C In the last step, we have x/2 = sin θ = opposite / hypotenuse, so cos θ = x 2 (draw a triangle). cosh(x) (g) sinh 2 (x) dx SOLUTION: Let u = sinh(x) and du = cosh(x) dx. Then cosh(x) sinh 2 (x) dx = du u 2 = u + C = sinh(x) + C. (h) sin 7 (x) cos 2 (x) dx SOLUTION: Let u = cos(x), du = sin(x) dx. Then sin 7 x cos 2 x dx = sin(x)( cos 2 (x)) cos 2 (x) dx = ( u 2 ) u 2 du = u 2 u + u u 8 du = u + u5 5 u7 7 + u9 9 + C = cos(x) + cos(x)5 5 cos(x)7 7 + cos(x)9 9 + C (i) x 2 x 2 dx SOLUTION: Let x = sec(θ), dx = tan θ sec θ dθ. Then x 2 x 2 dx = sec 2 (θ) tan θ sec θ dθ sec 2 θ = sec θ dθ

4 Now integrate by parts, with u = sec θ, du = sec θ tan θ dθ and v = tan θ, dv = sec 2 θ dθ. ( ) sec θ dθ = sec θ tan θ tan 2 θ sec θ dθ ( ) = sec θ dθ = sec θ tan θ (sec 2 θ ) sec θ dθ = = = Therefore, the final answer is sec (θ) dθ = sec θ tan θ + sec θ dθ sec θ dθ sec (θ) dθ = sec θ tan θ + ln tan θ + sec θ + C sec θ dθ = sec θ 2 + ln tan θ + sec θ + C. 2 x ln x 2 + x + C. cosh(x) (j) sinh(x) + dx SOLUTION: Let u = sinh(x) +, du = cosh(x) dx. Then cosh(x) sinh(x) + dx = 2 u du = ln u + C = ln sinh(x) + + C. x 2 + x (k) (x )(x + ) 2 dx SOLUTION: Perform partial fractions decomposition Clear denominators first. x 2 + x (x )(x + ) 2 = A x + B x + + Cx + D (x + ) 2 x 2 + x = A(x + ) 2 + B(x )(x + ) + (Cx + D)(x ) Plug in x = to get A =. Plug in x = to get 5 = D C. Plug in x = to get D = B. Plug in x = 2 to get = B + 2C + D. Substitute the previous equations to get B =, C = 5, and D =. So x 2 + x (x )(x + ) 2 dx = x + x + 5x (x + ) 2 dx = ln x + ln x + 5 x (x + ) 2 = ln x + ln x + 5 ln x + 5 x + + C (l) x 2 2 x dx

5 SOLUTION: Do long division. Divide x 2 2 by x to get x 2 2 x dx = (x + 2) + x2 dx = + 2x + ln x + C. x 2 (m) coth 2 ( t) dt SOLUTION: Let u = t, du = dt. Then coth 2 ( t) dt = coth 2 (u) du = csch 2 (u) + du = ( coth(u) + u + C) = coth( t) ( t) + C. (n) x 2 + x 5 dx SOLUTION: Perform partial fractions decomposition: x 2 + x 5 = A x B x + = = A(x ) + B(x + 5) Set x = = B = /. Set x = 5 = A = /. Then x 2 + x 5 = x + 5 x dx = (ln x + 5 ln x + C) (2) Find the volume of the solid obtained by rotating y = x x 2 about the y-axis. SOLUTION: Using the cylindrical shells method: V = = 2π ( ) 2πx x x 2 dx x 2 x 2 dx Let x = sin θ, dx = cos θ dθ. Then π/2 V = 2π sin 2 θ cos θ dθ Let u = sin θ, du = cos θ dθ. Then V = 2π u 2 du = 2π u = 2 π.

6 () Find the arc length of the graph of y = tan(x) over the interval [, π / ]. SOLUTION: Plug in to the arc length formula. π/ + (y ) 2 dx = π/ + sec (x) dx. No antiderivative exists, so numerical integration techniques must be used. () Suppose that a random variable X is distributed with density p(x) = C x 2 on [, ]. Find C such that p(x) defines a probability density function, and compute P( / 2 X ). SOLUTION: To find C, set = C x 2 dx Then evaluate the integral on the right and solve for C. To evaluate the integral, let x = sin θ, dx = cos θ dθ. Then π/2 C x 2 dx = C Therefore, C = 2/π. = C π/2 = C π 2 + C 2 cos 2 θ dθ 2 ( + cos(2θ)) dθ = C 2 ( ) π/2 2 sin(2θ) = C π 2 + C (sin π sin( π)) = C π 2. π/2 dθ + C 2 π/2 cos(2θ) dθ Now to find the probability, we ve already computed the antiderivative of p(x), so we just need to substitute new bounds. The probability is P( /2 X ) = /2 ( 2 x π 2 dx = 2 sin (x) + x ) x 2 π 2 /2 = 2 + π (5) Find C such that p(x) = Ce x e e x is a probability density function on (, ). SOLUTION: This is one of the homework questions from last week (also it was on the quiz). Go look at the homework solutions on blackboard. The answer is C =. () Suppose that a random variable X is distributed with density p(x) = x 2 e x2 on (, ). Find the mean of X. SOLUTION: µ = xp(x) dx = x e x2 dx This is an odd function integrated over a symmetric domain, so the integral is zero.

7 (7) Suppose that a random variable X is distributed with density p(x) = r e x/r on (, ). Find the mean of X. SOLUTION: This was on the homework last week, and I also did it in class. Go see the homework solutions on blackboard. The answer is µ = r. (8) Calculate T for the integral I = 2 x dx. SOLUTION: x = (2 )/ =. This is tedious, but easy to do. The answer is T = 27 (a) Is T too large or too small? Explain graphically. SOLUTION: Between x = and x = 2, the graph of y = x is concave up, so trapezoid rule overestimates the area under the graph; the trapezoids are above the graph. (b) Show that K 2 = f (2) may be used in the error bound and find a bound for the error. SOLUTION: K 2 is the max value of f (x) on the interval [, 2]. f (x) = x, so f (x) = x. Therefore, the maximum value of f (x) on the interval [, 2] happens at x = 2, and K 2 = f (2) = 2. Finally, Error K 2(b a) 2(2 ) n 2 = ( 2 = 2 ) 2 = 9. (c) Evaluate I and check that the actual error is less than the bound computed in (b). SOLUTION: This is easy to integrate. So the actual error is 2 x dx = x 2 = Error = 27 = 9. This is less than the error bound, which says that the error is at most /9. (9) Radium-22 has a half-life of 59 years. Consider a mass of mg of Radium-22. (a) What is the mass of Radium remaining after years? SOLUTION: The equation for radioactive decay is exponential decay, m(t) = m e t/t where m is the inital mass, m = mg, m(t) is the number of milligrams remaining after t years, and T is the half life, T = 5( years. Then m() = e /

8 (b) When will the mass of Radium be mg? SOLUTION: We want to know for which t does m(t) = mg. So set = e t/59 = = t 59 = t = 59 ln(/). () Show that e x2 dx converges using the Comparison Theorem. SOLUTION: The comparison theorem says that f (x) g(x) = f (x) dx g(x) dx. On the interval [, ), e x2 e x. Therefore, Hence, the integral converges. e x2 dx e x dx = e x =.