(x 3)(x + 5) = (x 3)(x 1) = x + 5. sin 2 x e ax bx 1 = 1 2. lim

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1 SMT Calculus Test Solutions February, x + x 5 Compute x x x + Answer: Solution: Note that x + x 5 x x + x )x + 5) = x )x ) = x + 5 x x + 5 Then x x = + 5 = Compute all real values of b such that, for fx) = x + bx 7, f) = f ) Answer: Solution: We have that f) = b and f ) = ) + b = b + 8 Setting these equal to each other, we see that b = Suppose a and b are real numbers such that x Determine all possible ordered pairs a, b) Answer:, ) and, ) sin x e ax bx = Solution: Since this is in an indeterminate form, we can use L Hôpital s Rule to obtain x sin x ae ax b = However, the numerator goes to zero, so the denominator must also go to zero to give us another indeterminate form This implies that a = b Using L Hôpital s Rule again, we have that cos x x a e ax = The numerator goes to, so the denominator must go to Therefore, a = b = ±, giving us a, b) =, ) and, ) Evaluate e x dx Answer: e + Let w = x so that w = x and dx = w dw Then the integral becomes To find this integral, use integration by parts: u = w du = dw; dv = e w dw v = e w we w dw = uv v du = we w e w dw = w )e w we w dw Evaluating w )e w at our its of integration yields e +

2 SMT Calculus Test Solutions February, sin 5x) tan x) 5 Evaluate x logx + )) 5 Answer: 5 Solution : For any function f with f) =, we know that fx) x x = fx) f) = f ) x x sin5x), tanx), and logx + ) are all at x =, and their derivatives at are 5,, and, respectively So, divide numerator and denominator by x 5 and re-arrange to get ) ) sin5x) sin 5x) tan x) x tanx) x x logx + )) 5 = ) x 5 = 5 5 = 5 logx+) x Solution : Recall from Taylor series that if f) =, then fx) f )x when x is small This allows us to write sin 5x) tan x) 5x) x) x logx + )) 5 = x x) 5 = 5 6 Compute k= Answer: π sin k x dx Bring the sum into the integral, so we have π sin k x dx k= The integrand is a geometric series, so the answer is π sin x dx = π π ) sec x dx = tan tan) = 7 The function fx) has the property that, for some real positive constant C, the expression f n) x) n + x + C is independent of n for all nonnegative integers n, provided that n + x + C Given that f ) = and fx) dx = C + e ), determine the value of C Note: f n) x) is the n-th derivative of fx), and f ) x) is defined to be fx) Answer: e Solution: Since f n) x)/n + x + C) is independent of n, we can say that it is equal to gx) Multiplying by n + x + C), we have that f n) x) = n + x + C)gx)

3 SMT Calculus Test Solutions February, Taking a derivative with respect to x, we obtain f n+) x) = n + x + C)g x) + gx) However, this is equal to n + + x + C)gx) by the problem statement Canceling terms, we obtain that gx) = g x) The only class of functions that is its own derivative is ae x, so we have that gx) = ae x for some constant a) Now, f x) = x + C + )ae x, so f ) = gives us that a = /C + ) We also have that Integration by parts gives us which simplifies to fx) dx = e )C + C + x + C C + ex dx = C + e ) = C + e ), C = e, from which it follows that the answer is e 8 The function fx) is defined for all x and is always nonnegative It has the additional property that if any line is drawn from the origin with any positive slope m, it intersects the graph y = fx) at precisely one point, which is m units from the origin Let a be the unique real number for which f takes on its maximum value at x = a you may assume that such an a exists) Find a fx) dx Answer: +log) Solution : First, express x and y as functions parametrized by m We have the system y = mx x + y = m Solving for y, we get y = m m Hence, maximizing y is equivalent to maximizing By +m +m differentiating with respect to m, we see that the maximum occurs when m =, at the point, ) Now, we just need to compute the integral However, this parametric form is not convenient Instead, by drawing the line y = x, we notice that the integral splits up into a right isosceles triangle, and a region between the line y = x and the y-axis This suggests that we should convert to polar coordinates In fact, fx) is equivalent to the graph rθ) =, since a line tanθ) at angle θ to the x-axis has slope tanθ) The area we wish to compute is π/ rθ) dθ = π/ cotθ) dθ = [logsinθ))]π/ π/ = log/ )) = log)

4 SMT Calculus Test Solutions February, ) We add this area to the area of the triangle, which is =, so our final answer is + log) Solution : We begin as before to find a, but present a different method of computing the integral Solving for x in terms of y, we get that x + y = x/y = yx x + y = = x = ± y y We only care about the region where ± y = xy, since x, y Hence, we take x = y y Notice that we can compute the desired quantity as ) y dy, y since within the square bounded by the coordinate axes and x, y, the area between the curve and the x-axis plus the area between the curve and the y-axis sum to the area of the whole square Now, using the substitution u = y, we get y dy = y = = = u)u u du u)u u) + u) du u + u du + u du = [u log + u)] = log) The answer is minus this quantity, so report + log) 9 Evaluate Answer: / dx sin x + cos x ) Solution : Observe that by pulling a factor of cos x out of the denominator, we can write the given integral as dx π/ + tan x) cos x = sec x dx + tan x)

5 SMT Calculus Test Solutions February, We now substitute u = tan x + : Thus, our integral is equal to which simplifies to du = sec x tan x dx = sec x u ) dx u u du = [ u + u ] u u du, = Solution : Let I be the value of the given integral Note that I = sinx) + cosx) ) ) dx, sinθ) ) which is the polar area bounded by the curve rθ) = + cosθ) and the x and y axes for θ [, π/] Converting to Cartesian coordinates, we get Therefore, Evaluate [ n Answer: π π π k= ) = r sinθ) + cosθ) ) = r sinθ) + r cosθ) = x + y = = y = x) = + x x I = = + x x dx [x + x x/ = + = 6 = I = ) ] k cos x) k x dx Solution : Observe that cos x) looks like a bunch of spikes, centered at, π, π,, each with area I n = cos x) dx ]

6 SMT Calculus Test Solutions February, We can integrate by parts to see that I k = = k ) cos x) k dx = [ ] π/ cos x) k sin x + k ) cos x) k cos x) dx = k )I k I k ) cos x) k sin x dx Therefore, I k = k k I n k = I n = k= ) k k I = π n k= k k As n, the spikes get sharper and sharper; this means that the denominator x of the integrand gets concentrated at x =, π, π, Therefore, we expect that as n, n ) k cos x) n ) k I n k x dx k kπ = π π = π π π k= k= k= Solution : We present a more rigorous approach here First, rewrite the problem into the following form: For each positive integer n, let a n = x dx Additionally, let c = n n k ), which is a positive finite constant Evaluate c a n k= Let B = {, π, π, } The idea is that the numerator of the integrand approaches a function with cπ area concentrated infinitely closely to each point in B Therefore, the it should be x dx = cπ π = cπ x π, x B where the last equality follows by the formula for summing geometric series We will soon get to a more precise way of thinking about the area being concentrated infinitely closely to points in B, but first let s see why the numerator should have cπ area around each point in B Since the area around each point in B is the same cos is periodic), we need only consider the area around We can apply integration by parts to find a formula for the area around in each term of the sequence The recurrence is Repeatedly applying this formula, we get dx = ) n ) dx dx = π n n k= ) k Taking the it as n, the area around goes to cπ So the answer makes sense Now we will prove it more rigorously

7 SMT Calculus Test Solutions February, For x / B, the integrand goes to as n because cos x < So for any open set x S sufficiently disjoint from B, we might guess that S x dx = If we require that every point in S is at least ɛ > away from any point in B, then this is indeed true There are two ways to see this The fanciest way to see it is to use the dominated convergence theorem, which says that if a sequence of functions f n converges pointwise to a funciton f and if there is some function ϕ with f n x) < ϕx) for all x S and S ϕ <, then S f n = S f To apply this theorem, we let f n be the integrand of the n-th term of the sequence To construct ϕ, notice that since every point in S is at least ɛ away from every point in B, there is some δ < so that cos x < δ for all x S So is bounded by nδ for all x S Since nδ has a finite it as n, is bounded by some finite number B for all x S So we can let ϕx) = B/ x Then f n x) < ϕx) for all x S and S ϕ <, just as we need in order to apply the theorem So we apply the theorem to get S x dx = S x dx = dx = S But we of course don t expect you to know the dominated convergence theorem, so we can also prove this using a bare hands method that is actually easier Bare hands is usually much harder than the dominated convergence theorem proof That is why people use the dominated convergence theorem But we have arranged for this problem to work with bare hands) As we argued above, there is some δ < so that cos x < δ for all x S Then < nδ for all x S So we have a bound S x dx nδ S x dx Since the integral converges, this goes to as n so we again have the desired result x The upshot of all this is that we can now define B ɛ to be the points in [, ) that are within ɛ of B and have x dx = B ɛ x dx To calculate the integral on the right, notice that it is just the sum over all integers k of kπ+ɛ kπ ɛ x dx = ɛ kπ ɛ x dx By the formula for summing geometric series, the sum of this over all integers k is ɛ x dx = π ncos x) π ɛ x dx ) So we have reduced the problem to calculating the following it: ɛ ɛ x dx

8 SMT Calculus Test Solutions February, To do this, bound the it above and below by taking the highest and lowest possible values of x out of the integral: ɛ ɛ ɛ ncos x) ɛ dx ɛ ɛ x dx ɛ ncos x) dx ɛ By a very similar argument as above, nothing outside ɛ, ɛ) contributes to the integrals in our bounds and therefore ɛ π/ ncos x) dx = ncos x) dx ɛ We have already calculated the right hand side: it is cπ So we can plug this back into our bounds to get ɛ ɛ cπ ɛ x dx ɛ cπ Plugging this bound into ) gives ɛ cπ π ncos x) π x dx ɛ cπ π π Since ɛ was arbitrary, taking ɛ forces x dx = cπ π π, as desired Finally, you might be interested in knowing why c is a positive finite constant This is not necessary to solve the problem, but it is necessary to be sure that the problem makes sense And it is interesting) To see this, let b n = n n k= k ) Then log b n = n log n + log ) = n k log n + )) k + O k k= k= Here O ) denotes some function of k whose absolute value is always less than C for some C k k big enough The fact that log k ) = k + O ) follows from the Taylor expansion of log k around It is well known that n k= k = log n + γ + O n ) where γ is some constant Plugging this in gives log b n = log n log n γ + O n ) + n k= ) O k = ) γ + O + n n k= ) O k Since k= converges to some constant, n k k= O ) converges to some finite constant α as k n Therefore log b n γ+α as n This is some finite number, so c = exp γ+α) is a positive finite number, as desired