Category Archives: More Challenging Problems

More Challenging Problems: Numerical Sequences and Series

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Introductory Problems

1. Suppose a1 = 1 and for all n > 2, an = (an-1 + 5)/2. Show the sequence an converges and find    the limit to which it converges.

Answer

1. The sequence is increasing and bounded above by 5, so it converges. The limit of the terms is 5

Solution

1. Each term after a1 is the midpoint of 5 and the previous term, so the sequence is increasing and bounded above by 5. It follows from the Monotone Convergence Theorem that this sequence converges. Now that we know the limit limn → ∞an exists, denote this limit by α. Taking the n → ∞ limit of both sides of the equation: an= (an-1 + 5)/2 we find: α= (α + 5)/2 Solving for α gives α= 5.

 

2. Suppose a1 = 1, a2 = 1, and for all n > 2, an= an-1 + an-2. Give numerical evidence that the sequence bn = an/an+1 converges. Assuming the sequence converges, find the limit to which it converges.

Answer

1. Convergence follows from two applications of the Monotone Convergence Theorem, and an estimate on the difference of successive terms. The sequence converges to (-1 + √5)/2.

Solution

1. The first few values of the ratios bn= an/an+1 are 1, .5, .6667, .6000, .6250, .6154, .6190, .6176, .6182, .6180, .6181, .6180, .6180, … . It appears that the sequence does converge. Now assuming the limit limn → ∞an/an+1 exists, denote this limit by α. Divide both sides of the equation an= an-1 + an-2 by an-1, obtaining an/an-1= 1 + an-2/an-1 Taking the n → ∞ limit of both sides of this equation gives 1/α= 1 + α Multiplying both sides by α gives a quadratic equation with solutions α= (-1 ± √5)/2. Being the limit of ratios of positive terms, α cannot be negative, so α= (-1 + √5)/2

 

3. Does the series ∑n=1∞an with an = log(n/(n+1)) converge or diverge?

Answer

1. The series diverges.

Solution

1. This is easy if you recall the quotient property of logarithms log(n/(n+1)) = log(n) – log(n+1) So the series telescopes (log(1) – log(2)) + (log(2) – log(3)) + (log(3) – log(4)) + … so the nth partial sum is Sn = log(1) – log(n+1) = -log(n+1), which diverges to -∞.

 

4. Suppose a1= 1/2 and for all n > 1, an= (1/2)n + an-1. Does the series a1 + a2 + a3 + … converge or diverge?

Answer

1. The series diverges.

Solution

1. Let’s look for a pattern in the first few terms: a1= 1/2 a2= 1/22 + 1/2 = (1 + 2)/22 a3= 1/23 + (1 + 2)/22 = (1 + 2 + 22)/23 and in general an = (1 + 2 + … + 2n)/2n+1 Now recall 1 + 2 + 22 + … + 2n= 2n+1 – 1. (Easy to prove by induction if you don’t recall this result.) So we see that: an= (2n+1 – 1)/2n+1 and the limn → ∞an= 1. The series diverges by the nth term test.

 

 

 

Numerical Sequences and Series: The ratio and root tests

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More Challenging Problems

Determine in the series a1 + a2 + a3 + … converges or diverges.

 

1. an = n2/3n

Answer

1. The series converges.

Solution

1. Use the Ratio test:                                                                                                                                   limn → ∞an+1/an                                                                                                                                                                                                = limn → ∞((n+1)2/3n+1)/ (n2/3n)                                                                                                                    = limn → ∞((n+1)2/n2)⋅(3n/3n+1)                                                                                                                      = limn → ∞((n+1)/n)2⋅(1/3)= 1/3                                                                                                                   So the series converges by the ratio test.

 

2. an= 3n/nn

Answer

1. The series converges.

Solution

1. The exponent n suggests the Root Test:                                                                                           limn → ∞(an)1/n                                                                                                                                                = limn → ∞(3n/nn)1/n                                                                                                                                           = limn → ∞ 3/n = 0                                                                                                                                           So the series converges by the Root Test.

 

3. an = n/(ln(n)n)

Answer

1. The series converges.

Solution

1. The exponent n in the denominator suggests the Root Test:                                                           limn → ∞(an)1/n                                                                                                                                             = limn → ∞(n/(ln(n)n))1/n                                                                                                                                    = limn → ∞ (n1/n)/ln(n)                                                                                                                                     For the n1/n factor, write y = x1/x, so ln(y) = ln(x)/x. Applying l’Hôspital’s rule, we see ln(y) → 0, so y → 1. Then with the ln(n) in the denominator, we see limn → ∞(an)1/n= 0, so the series converges by the Root Test.

 

4. an = (2n/(n + 1))n

Answer

1. The series diverges.

Solution

1. The exponent n suggests the Root Test:                                                                                           limn → ∞(an)1/n                                                                                                                                                   = limn → ∞((2n/(n + 1))n)1/n                                                                                                                             = limn → ∞ 2n/(n + 1) = 2                                                                                                                              So the series diverges by the Root Test.

 

5. an= n!/en

Answer

1. The series diverges.

Solution

1. he factorial suggests we try the Ratio Test:                                                                                       limn → ∞an+1/an                                                                                                                                                                                                = limn → ∞((n+1)!/en+1)/ (n!/en)                                                                                                                      = limn → ∞((n+1)!/n!)⋅(en/en+1)                                                                                                                        = limn → ∞ (n+1)/e So the ratio an+1/an → ∞ as n → ∞ and we see the series diverges by the Ratio Test.

 

6. an = n!/nn

Answer

1. The series converges.

Solution

1. Here we have both a factorial and an exponent n. In case you were thinking that factorial always means Ratio Test and that exponent n always means Root Test, this problem shows that rule cannot be applied to all series. Our intuition suggests we won’t have much luck with roots of factorials, but maybe we can handle ratios of exponents. So let’s try the Ratio Test:                 limn → ∞an+1/an                                                                                                                                                = limn → ∞((n+1)!/(n+1)n+1) / (n!/nn)                                                                                                               = limn → ∞((n+1)!/n!)⋅(nn/(n+1)n+1)                                                                                                                 = limn → ∞(n+1)⋅(nn/(n+1)n+1)                                                                                                                        = limn → ∞nn/(n+1)n                                                                                                                                        = limn → ∞(n/(n+1))n                                                                                                                              Recalling that: limn → ∞(1 + 1/n)n = we see: limn → ∞an+1/an= 1/e < 1 and so the series converges by the Ratio Test.

 

 

More Challenging Problems: Complex numbers and Euler’s formula

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Introductory Problems

1. Solve z2 + z = 1 + i

Answer

1. z = i and z = -1 – i

Solution

1. Apply the quadratic formula to z2 + z + 1 – i = 0, obtaining z= (-1 ± (-3 + 4i)1/2)/2 To compute the square root, we find the polar representation of -3 + 4i. The modulus is ((-3)2 + 42)1/2= 5. The argument is θ = arctan(-4/3). Then (-3 + 4i)2 has modulus √5 and argument θ/2. This gives: z= (-1 ± √5(cos(θ/2) + i sin(θ/2)/2 = (-1 ± √5(cos(θ/2)) + i √5 sin(θ/2)/2. But we can do better. Using the half-angle formulas: cos(θ/2) = ((1 + cos(θ))/2)1/2 and sin(θ/2) = ((1 – cos(θ))/2)1/2 we need to find cos(θ) when tan(θ) = -3/4. Recalling we started with the polar representation of -3 + 4i, we see cos(θ)= -3/5.

 

2. Find Cartesian expressions for (a) ei and (b) e1+i.

Answer

1 The answers are the following: (a) ei = cos(1) + i sin(1) (b) e1+i = e⋅cos(1) + i e⋅sin(1)

Solution

1. The solutions are as follows: (a) Apply Euler’s formula with z = 1: ei = ei1 = cos(1) + i sin(1). (b) e1 + i = e1ei = e(cos(1) + i sin(1)).

 

3. Find Cartesian expressions for (a) cos(i) and (b) cos(1+i).

Answer

1. Th answers are as follows: (a) (e1 + e-1)/2 (b) cos(1 + i) = cos(1)(e1 + e-1)/2 + i sin(1)(e1 – e-1)/2

Solution

1. The solutions are as follows: (a) From the power series for cos(z) we see cos(i) = 1 – i2/2! + i4/4! – i6/6! + i8/8! – … = 1 + 1/2! + 1/4! + 1/6! + 1/8! + … (e1 + e-1)2 (b) Use the angle sum formula for cosine to find cos(1 + i) = cos(1)cos(i) + sin(1)sin(i). We know all the parts of this except sin(i), which we find from its power series sin(i) = i – i3/3! + i5/5! – i7/7! + … = i(1 + 1/3! + 1/5! + 1/6! + … ) = i(e1 – e-1)/2 Combining these, we see cos(1 + i) = cos(1)(e1 + e-1)/2 + i sin(1)(e1 – e-1)/2.

 

4. Describe the curves: (a) ex+ iπ/2 and ex + iπ for all x (b) e0 + iy and e1 + iy for all y

Answer

1. The answers are the following: (a) ex + iπ/2 is the positive imaginary axis; ex + iπ is the negative real axis. (b) e0+ iy is the circle of radius 1 centered at the origin; e01+ iy is the circle of radius e centered at the origin

Solutions

1. We use ex +yi= ex(cos(y) + i sin(y)). (a) ex + iπ/2 = ex(cos(π/2) + i sin(π/2)) = exi, the positive imaginary axis. ex + iπ= ex(cos(π) + i sin(π))= ex(-1), the negative real axis. (b) e0 + iy = e0(cos(y) + i sin(y)) = cos(y) + i sin(y), the circle of radius 1 centered at the origin. e1 + iy = e1(cos(y) + i sin(y)) = the circle of radius e centered at the origin.

 

5. Find the period of the complex cosine function.

Answer

1. 2π

Solution

1. By the angle sum formula sin(a + b) = sin(a)cos(b) + cos(a)sin(b) we see sin(z + π/2) = cos(z). From this we see that sin(z) and cos(z) have the same period, call it α. Then by Euler’s formula, ei(z + α)= cos(z + α) + isin(z + α) = cos(z) + i sin(z) = eiz From this we see: ez + iα= ei(-iz + α)= ei(-iz)= ez That is, ez is periodic with period iα, so α= 2 π.

 

 

 

 

 

More Challenging Problems: Area of surfaces of revolution

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Introductory Problems

 

1. Show that the surface obtained by revolving x = g(y), c ≤ y ≤ d, about the y-axis has area ∫cd 2π g(y) √(1 + (g ‘(y)2) dy

Solution

1. Rotating x= g(y) around the y-axis gives a circle of radius g(y), so the circumference 2πf(x) in the formula for revolving about the x-axis is replaced by the circumference 2π g(y). Similarly, the arclength factor √(1 + (f ‘(x)2) is replaced by the arclength factor √(1 + (g ‘(y)2). The result follows by combining these factors.

 

2. Find the area of the surface obtained by revolving x= y3, 0 ≤ y ≤ 1, about the y-axis.

Answer

1. The area is (π/27)(10√10 – 1)

Solution

1. Because g(y)= y3, g ‘(y) = 3y2 and the area is: ∫012π y3 √(1 + (3y2)2) dy = 2π∫01y3√(1 + 9y4) dy = 2π(1/36)(2/3)u3/2|110 by substituting u= 1 + 9y4 = (π/27)(10√10 – 1)

 

3. For the surface obtained by revolving x = y3, 0 ≤ y ≤ 2, about the y-axis, show the area is greater than 64π/5.

Solution

1. Because g(y) = y4, g ‘(y) = 4y3 and the area is ∫022π y4 √(1 + (4y3)2) dy = 2π∫02 y4 √(1 + 16y6) dy No obvious – or non-obvious, as far as we can tell – trick works to evaluate this integral. But notice that the terms inside the square root are always ≥ 1, so area ≥ 2π∫02 y4 dy = 64π/5.

 

4. For the surface obtained by revolving x= 1/y, 1 ≤ y ≤ 2, about the y-axis, show the area is greater than 2πln(2).

Solution

1. Because g(y) = 1/y, g'(y)= -1/y2 and the area is: ∫122π 1/y √(1 + (-1/y2)2) dy = 2π∫12 1/y √(1 + 1/y4) dy No obvious – or non-obvious, as far as we can tell – trick works to evaluate this integral. But notice that the terms inside the square root are always ≥ 1, so area ≥ 2π∫121/y dy = 2π ln(y)|12 = 2π ln(2).

 

5. Find the value of a for which the surface obtained by revolving y= x3, 0 ≤ x ≤ a, about the x-axis has area 2π.

Answer

1. The value of a= ((552/3 – 1)/9)1/4 ≈ 1.106.

Solution

1. Because f(x)= x3, f'(x)= 3x2 and the area is: ∫0a2π x3 √(1 + 9x4) dx = 2π((1 + 9a4)3/2 – 1)/54 Setting area = 2π and solving for a gives a = ((552/3 – 1)/9)1/4.

 

 

 

 

 

 

More Challenging Problems: Arclength

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Introductory Problems

 

1. Find the length of f(x)= x3/6 + 1/(2x) from x = 1 to x = 2.

Answer

1. The arclength is 17/12.

Solutions

1. First, f'(x)= x2/2 – 1/(2x2), so √(1 + (f'(x)2) = √(x4/4 + 1/2 + 1/(4x4)) = √((x2/2 + 1/(2x2))2). Then the arclength is: ∫12x2/2 + 1/(2x2) dx = ((x3/6) – 1/(2x))|12 = 17/12.

 

2. Find the length of f(x) = x2/2 – ln(x)/4 from x = 1 to x = 2.

Answer

1. The arclength is 3/2 + ln(2)/4.

Solution

1. First, f'(x)= x – 1/(4x), so √(1 + (f'(x))2) = √(x2/4 + 1/2 + 1/(16x2) = √((x + 1/(4x)(2). Then the arclength is ∫12(x + 1/(4x)) dx = ((x2/2) – ln(x)/4)|12 = 3/2 + ln(2)/4.

 

3. Find the length of f(x)= ln(cos(x)) from x = 0 to x = π/4.

Answer

1. The arclength is ln(√2 + 1).

Solution

1. First, f'(x)= -sin(x)/cos(x) = -tan(x), so √(1 + (f'(x)2) = √(1 + tan2(x)) = √(sec2(x)) = sec(x), because sec(x) is positive in this range of x values. Then the arclength is: ∫0π/4sec(x) dx = ln|(sec(x) + tan(x)||0π/4(by 42 from the integral table) = ln(√2 + 1)

 

4. Find the length of r(t) = < et, (1/2)e2t> from t= 0 to t= 1

Answer

1. The arclength is ((e/2)√(1 + e2) + (1/2)ln(1 + √(1 + e2)) – ((√2)/2 + (1/2)ln(1 + √2)).

Solutions

1. First, r'(t)= < et, e2t > and |r'(t)| simplifies to et√(1 + e2t) Then the arclength is: ∫01et√(1 + e2t) dt= (e/2)√(1 + e2) + (1/2)ln(1 + √(1 + e2)) – ((√2)/2 + (1/2)ln(1 + √2)), where we substituted x = et and applied 16 from the integral table.

 

5. Find the length of f(x) = ln(x) from x = 1 to x = 2.

Answer

1. The arclength is (√5 – ln((1 + √5)/2)) – (√2 – ln(1 + √2)).

Solution

1. First, f'(x)= 1/x, so √(1 + (f'(x)2) = √(1 + 1/x2) = (√(x2 + 1))/x. Applying formula 18 from the integral table, we find the arclength is: (√5 – ln((1 + √5)/2)) – (√2 – ln(1 + √2)).

 

 

 

 

More Challenging Problems: Parametric curves

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Introductory Problems

1. Plot x(t) = cos(t), y(t) = cos(kt) for k = 1, 2, and 3. Explain why the graphs have these shapes. (Hint: think of trigonometric identities.)

Answer

1. Here are the plots for k = 1, 2, and 3.

 

k=1                                                  k=2                                                    k=3

Solution

1. The k= 1 case is easy: y(t) = x(t) is the strainght line of slope 1 passing through the origin. For k = 2, use the trigonometric identity cos(2t) = 2cos2(t) -1 to obtain y(t) = 2x(t)2 – 1, a parabola opening upward with minimum at (0, -1). For k = 3, use cos(a + b) = cos(a)cos(b) – sin(a)sin(b) and sin(2a) = 2sin(a)cos(a) to obtain y(t) = cos(3t) = cos(t + 2t) = cos(t)cos(2t) – sin(t)sin(2t) = cos(t)(2cos2(t) – 1) – sin(t)2sin(t)cos(t) = 2cos3(t) – cos(t) – 2cos(t)(1 – cos2(t)) = 4cos3(t) – 3 cos(t) = 4x(t)3 – 3x(t)

 

2. Write the equation for the tangent line to the curve x(t) = t2, y(t) = t3 – t at the point (1,0).

Answer

1. The curve passes through (x, y)= (1, 0) twice, and has two tangent lines at that point: y= -(x – 1) and y= x – 1

Solution

1. The slope of the tangent line is: dy/dx = (t3 – t)’/(t2)’= (3t2 – 1)/(2t) To find the slope of the tangent line at the point (x, y) = (1, 0), we must find the value of t corresponding to that point. First, y= 0 means t3 – t= 1; that is, t= -1, t= 0, and t= 1. Second, x= 1 means t2= 1; that is, t= -1 and t= 1. Consequently, this curve passes through the point (x, y) = (1, 0) at t = -1 and at t = 1. At t = -1 the slope of the tangent line is -1 and so the equation of the tangent line is y = -(x – 1) At t = 1 the slope of the tangent line is 1 and so the equation of the tangent line is y = (x – 1)

 

3. Sketch the curve x(t) = t cos(1/t), y(t) = t sin(1/t), z(t) = (1 – t2)1/2 of 0 < t < 1. Hint: On what surface does it lie?

Answer

1. Please note the following image:

Solution

1. First note that x2 + y2 = t2, so x2 + y2 + z2= 1. That is, the curve lies on the unit sphere. In fact, on the upper hemisphere because the z-coordinate is positive. Projected into the xy-plane, the curve is a spiral converging to (0,0). The spiral makes a complete turn between 1/2π and 1/4π, between 1/4π and 1/6π, between 1/6π and 1/8π, and so on, spinning ever faster. Moreover, the distance to (0, 0) decreases as t decreases toward 0. Combining these observations, we see the curve is a spiral on the upper hemisphere of the unit sphere, spiraling in to the north pole.

 

4. Sketch the curve x(t) = t cos(1/t), y(t) = t, z(t) = t sin(1/t), 0 < t < 1. Hint: on what surface does this curve lie?

Answer

1. Please note the following image:

Solution

1. First, note that x2 + z2= t2, so the curve lies on the cone y = (x2 + z2)1/2, having the positive y-axis as its axis. Projected into the xz-plane, the curve is a spiral converging to (0,0), so in three dimensions, the curve spirals to the apex of the cone.

 

5. Plot the curve x(t)= t – sin(t), y(t)= 1 – cos(t), for 0 ≤ t ≤ 2π. Locate all horizontal and vertical tangents, and determine the concavity of the graph.

Answer

1. The curve has vertical tangents at (0,0) and at (2π,0), and has a horizontal tangent at (π,2). The curve is concave down everywhere.

 

Solutions

1. If the curve has a horizontal tangent at parameter t, then 0 = (1 – cos(t))’ = sin(t), so t = 0, π, and 2π are the possible parameters of horizontal tangents. If the curve has a vertical tangent at parameter t, then 0 = (t – sin(t))’ = 1 – cos(t), so t = 0 and 2π are the possible parameters of vertical tangents. First, we see that the curve has a horizontal tangent at t = π, at the point (x, y) = π,2). To check of the curve has a horizontal or vertical tangent, or neither, at t = 0 and t = 2π, compute the limit limt → 0dy/dx = limt → 0(dy/dt)/(dx/dt) = limt → 0(sin(t))/(1 – cos(t)) = limt → 0(cos(t))/(sin(t)) by l’Hopital’s rule = ∞ So the curve has a vertical tangent at (0,0). A similar calculation shows the curve has a vertical tangent at (2π, 0). To determine concavity, we compute d2y/dx2, a bit tricky because we do not know y as an explicit function of x. So instead we use the chain rule, twice. d2y/dx2 = (d/dx)(dy/dx) = (d/dx)((dy/dt)/(dx/dt)) = (d/dx)(sin(t)/(1 – cos(t))) = (d/dt)(sin(t)/(1 – cos(t))) / (dx/dt) = -1/(1 – cos(t)) The curve is concave down throughout its domain.

More Challenging Problems: Volumes by cross-section

Notes PDF
Introductory Problems

1. Find the volume of the solid with cross-section a rectangle of base x and height ex, 0 ≤ x ≤ 1.

Answer

1. Volume = 1.

Solutions

1. The area is A(x) = base ⋅ height = x⋅ex. The volume is V = ∫01A(x) dx = ∫01 x⋅ex dx= (x⋅ex – ex)|01 = 1 where x⋅ex was integrated by parts, using u= x and dv= exdx.

 

2. Find the volume of the solid obtained by rotating the curve y= x2, -1 ≤ x ≤ 1, about the line y= -2.

Answer

1. Volume = π⋅166/15.

Solution

1. Rotating the curve y = x2 about the line y= -2 gives cross-sections that are circles of radius 2 + x2

 

 

Then A(x) = π⋅(2 + x2)2 = π⋅(4 + 4×2 + x4), and so the volume is V = ∫-11π⋅(4 + 4×2 + x4) dx = π⋅(4x + (4/3)x3 + (1/5)x5)|-11 = π⋅166/15.

 

3. Find the volume of the solid obtained by rotating the curve y = 2x – 2x2, 0≤ x ≤1, about the line y = 1

Answer

1. Volume = π⋅7/15.

Solutions

1. Rotating the curve y = 2x – 2x2 about the line y = 1 gives cross-sections that are circles of radius 1 – (2x – 2x2)

Then A(x) = π⋅(1 – (2x – 2x2)2= π⋅(1 – 4x + 8x2 – 8x3 + 4x4), and so the volume is V = ∫01π⋅(1 – 4x + 8x2 – 8x3 + 4x4) dx = π⋅(x – 4(x2/2) + 8(x3/3) – 8(x4/4) + 4(x5/5)|01 = π⋅7/15.

 

4. The questions are the following:

(a) Find the volume of the solid obtained by rotating the region between y = 1 and y= x2, 0 ≤ x ≤ 1, about the x-axis.

(b) Find the volume of the solid obtained by rotating the region between y = 1 and y= x3, 0 ≤ x ≤ 1, about the x-axis.

(c) Find the volume of the solid obtained by rotating the region between y = 1 and y= xn, 0 ≤ x ≤ 1, about the x-axis. Comment on the volume as n → ∞. Does this make sense? (It does, so perhaps the question should be “Why does this make sense?”)

Answer

1. The answers are the following: (a) Volume = 4π/5. (b) Volume = 6π/7. (c) Volume = 2nπ/(2n+1).

Solution

1. (a) Rotating the region between y = 1 and y = x2 around the x-axis gives an annular cross-section, the region between a circle of radius 1 and a circle of radius x2.

 

Then A(x) = π⋅(12 – (x2)(2

– (x2)(2) = π⋅(1 – x4), and so the volume is V = ∫01π⋅(1 – x4) dx = π⋅(1 – x5/5)|01= 4π/5. (b) Here the radius of the inner circle is x3, so A(x) = π⋅(1 – (x3)2) and V = ∫01π⋅(1 – x6) dx = π⋅(1 – x7/7)|01 = 6π/7. (c) Now the radius of the inner circle is xn, so A(x) = π⋅(1 – (xn2) and V = ∫01π⋅(1 – x2n) dx = π⋅(1 – x2n+1/(2n+1))|01 = 2nπ/(2n+1). As n → ∞, volume → π. This is the volume of the cylinder obtained by rotating the line y = 1, 0 ≤ x ≤ 1 about the x-axis. This is reasonable, because as n → ∞ the curve y = xn looks more and more like the line from (0,0) to (1,0), followed by the line from (1,0) to (1,1). That is, the inner radius of the annulus approaches 0.

Suppose for 1 ≤ x ≤ 2, perpendicular to the x-axis a solid has rectangular cross-section with base x2. Find the height function h(x) = xa so the solid has volume 1

 

 

5. Suppose for 1 ≤ x ≤ 2, perpendicular to the x-axis a solid has rectangular cross-section with base x2. Find the height function h(x)= xa so the solid has volume 1.

Answer

1. The value a= -2 works.

Solutions

1. The base length is x2 and the height is xa, so the cross-sectional area is A(x) = base⋅altitude = x2⋅xa = xa+2. Then the volume is V = ∫12xa+2 dx = (xa+3)/(a + 3)|12= (2a+3 – 1)/(a + 3). Then setting volume = 1 gives 2a+3= 4 + a. This can be solved numerically, or by plotting y = 2x+3 and y = 4 + x on the same graph. Or in this case we can guess, and find a = -2 works.

 

 

More Challenging Problems: Using integration tables

Notes PDF
Introductory Problems

 

1. Use formula 8. in the notes to find ∫cos2(x) dx. Compare your answer with formula 2

Answer

1. 1/2 cos(x) sin(x) + x/2 + C

Solution

1. Using formula 8. from the notes, with n= 2, tells us that ∫cos2(x) dx = 1/2 cos(x) sin(x) + 1/2 ∫ cos0(x) dx Since cos0(x) = 1, the integral on the right side evaluates to: ∫cos0(x) dx = ∫dx= x + c We can put that back in our original integral, and get: ∫cos2(x) dx = 1/2 cos(x) sin(x) + x/2 + C The answer given by formula 2 is: ∫cos2(x) dx = x/2 + 1/4 sin(2x) + C If we use the identity sin(2x) = 2 sin(x) cos (x), we see that the results are the same.

 

2. Find ∫cos4(x) dx

Answer

1. 1/4 cos3(x) sin(x) + 3/8 cos(x) sin(x) + 3/8 x + C

Solution

1. We can use formula 8. from the notes with n = 4, to reduce the power of cos(x) in the expression: ∫cos4x dx = 1/4 cos3(x) sin(x) + 3/4 ∫cos2(x) dx To evaluate the integral on the right side, we can use our answer to the previous problem (or, alternatively, formula 2. in the notes). Our answer was ∫cos2(x) dx = 1/2 cos(x) sin(x) + x/2 + C Putting that in, we get: ∫cos4x dx = 1/4 cos3(x) sin(x) + 3/8 cos(x) sin(x) + 3/8 x + C

 

3. Find ∫1/x [1 – ln2(x)]-1/2 dx

Answer

1. Arcsin [ln(x)] + c

Solution

1. This looks like formula 30. from the notes, except we have ln2(x) under the square root instead of x2. Let’s try using substitution first, with u = ln2(x). Then du = 1/x dx, and we get ∫1/x [1 – ln2(x)]-1/2 dx = ∫[1 – u2]-1/2 du The factor 1/x disappeared, and we’re in the territory of formula 28. Using it gives us ∫1/x [1 – ln2(x)]-1/2 dx = ∫[1 – u2=>]-1/2 du = arcsin(u) + c = arcsin [ln(x)] + c

 

4. Find ∫cos(x) [1 – sin2(x)]-1/2 dx

Answer

1. ± x + C

Solution

1. This looks very much like formula 28. in the notes, except it’s for sin(x) instead of x. But we have cos(x) up front, which is the derivative of sin(x): perfect case for substitution u = sin(x), du = cos(x) dx, followed by formula 28. for u: ∫cos(x) [1 – sin2(x)]-1/2 dx = ∫[1 – u2]-1/2du = arcsin(u) + c = arcsin [sin(x)] + c This is interesting. We are getting arcsin[sin(x)] + c = ± x + nπ + c The nπ + c can be replaced by a new constant C, and we get, finally, that ∫cos(x) [1 – sin2(x)]-1/2 dx = ± x + C That seems way too easy for such a complicated expression. We could have simplified it right at the beginning, by noting that 1 – sin2(x) = cos2(x). The integral is ∫cos(x) [1 – sin2(x)]-1/2 dx = ∫cos(x)/| cos(x) | dx= ∫± dx = ± x + C Even with tables at hand, it pays to check for simplifications first, before looking anything up.

 

5. Use integration by parts and formula 24. from the notes to find ∫x2 [1 – x2]-1/2 dx. Compare your answer to formula 29

Answer

1. -x/2 [1 – x2]1/2 + 1 /2 arcsin(x) + c

Solution

1. What to take for dv? We know how to integrate both x and x2, but setting either to dv will make the situation worse, as we have seen before in similar problems: the power of x will increase, and that’s not helpful in this case. With the hint to use formula 28., we may as well set dv = [1 – x2]-1/2, which gives us the variables as follows: The easiest factor to integrate is dv = x[1 – x2]-1/2, using substitution with u= x2: ∫x[1 – x2]-1/2 = -1/2 [1 – x2]1/2 + C With this, our substitution variables are: dv = x[1 – x2]-1/2, u = x, so du = 1, v = -[1 – x2]1/2 Integrating by parts, we get: ∫x2[1 – x2]-1/2 dx = – x [1 – x2]1/2 + ∫[1 – x2]1/2 dx We can use formula 24. for the right hand side integral: ∫[1 – x2]1/2 dx = + x / 2 [1 – x2]1/2 + 1 /2 arcsin(x) + c Putting this into our original answer gives: ∫x2 [1 – x2]-1/2 dx = -x [1 – x2]1/2 + x/2 [1 – x2]1/2 + 1 /2 arcsin(x) + c = -x / 2 [1 – x2]1/2 + 1 /2 arcsin(x) + c This is exactly the same as formula 29.

 

 

More Challenging Problems: Integration by parts

Notes PDF
Introductory Problems

 

1. Find ∫ln2(x) dx

Answer

1. x ln2(x) – 2x ln(x) + 2x + c

Solution

1. Recall that ∫ln(x) dx= x ln(x) – x. We can either take dv = 1 or dv = ln(x), since we know how to integrate both. The former is perhaps slightly easier. We have: dv= 1, u= ln2(x), so v= x, du= 2 ln(x) / x Now we integrate by parts: ∫ln2(x) dx= x ln2(x) – ∫2ln(x) dx = x ln2(x) – 2x ln(x) + 2x + c You can try starting with dv= ln(x) for practice, see if you can get the same answer.

 

2. Find ∫cos2(x) dx using integration by parts

Answer

1. 1/2 [sin(x) cos(x) + x + c]

Solution

1. Use u= dv= cos(x). Then du= – sin(x), v= sin(x) and we get ∫cos2(x) dx= sin(x) cos(x) + ∫sin2(x) dx Now we substitute [1 – cos2(x)] for the sin2(x) and the integral on the right side becomes ∫sin2(x) dx= ∫[1 – cos2(x)] dx = x – ∫cos2(x) dx We put this into the original equation, and gather up both cos2(x) terms to get ∫cos2(x) dx = 1/2 [sin(x) cos(x) + x + c]

 

3. Find ∫cos[ln(x)] dx

Answer

1. x/2 (cos[ln(x)] + sin[ln(x)]) + c

Solution

1. The only choice here is to set dv = 1: u = cos[ ln(x) ], dv = 1, so v = x and du = – sin[ ln(x) ] / x Integrating by parts: ∫ cos[ ln(x) ] dx = x cos[ ln(x) ] + ∫ sin[ ln(x) ] dx This does not seem to be of much help. We can try integrating ∫sin[ln(x)] dx by parts, to see if we can get to anything useful. Again, set dv = 1: u = sin[ln(x)], dv = 1, so v = x and du = cos[ln(x)]/x We get: ∫sin[ln(x)] dx = x sin[ln(x)] – ∫cos[ ln(x)] dx Altogether, we have ∫cos[ln(x)] dx = x cos[ln(x)] + x sin[ln(x)] – ∫cos[ ln(x)] dx We can put both integrals on the left, and arrive at the answer: ∫cos[ln(x)] dx = x/2 ( cos[ ln(x) ] + sin[ ln(x) ] ) + c

 

4. Find ∫ex cos(x) dx

Answer

1. 1/2 [ex sin(x) + ex cos(x)] + c

Solution

1. In this case, it is not at all clear which factor we should pick for u and which for dv. It turns out that the choices lead to a very similar solution. Let us choose dv = cos(x). u = ex, dv = cos(x), so du = ex, v = sin(x) Integrating by parts: ∫excos(x) dx= ex sin(x) – ∫ex sin(x) dx We integrate by parts again, choosing u= ex again (otherwise we would get back where we started): u = ex, dv= sin(x), so du= ex, v= – cos(x) We get: ∫ex sin(x) dx = -ex cos(x) + ∫ex cos(x) dx Altogether, we have: ∫ex cos(x) dx= ex sin(x) + ex cos(x) – ∫ex cos(x) dx Combining the integrals gives us the answer: ∫ex cos(x) dx = 1/2 [ex sin(x) + ex cos(x)] + c

 

5. Find ∫x (x-1)100 dx

Answer

1. 1/101 x (x-1)101 – 1/101 ⋅ 1/102 (x-1)102 + c

Solution

1. Let’s not multiply it out. Instead, use integration by parts, with u = x: u = x, dv = (x-1)100, so du = 1, v = 1/101 (x-1)101 We get: ∫x (x-1)100 dx = 1/101 x(x-1)101 – ∫1/101 (x-1)101 dx = 1/101 x (x-1)101 – 1/101 ⋅ 1/102 (x-1)102 + c

 

 

 

More Challenging Problems: Integration by substitution problems

Notes PDF
Introductory Problems

 

1. Find ∫ex/[1 + ex] dx

Answer

1. ln (1 + ex) + c

Solution

1. Substitute u= ex. Then du= ex dx, and we get ∫ex/[1 + ex] dx = ∫1/(1 + u) du = ln(1 + u) + c = ln (1 + ex) + c

 

2. Find ∫cos5(x) dx

Answer

1. sin(x) – 2/3 sin3(x) + 1/5 sin5(x)

Solution

1. We can write: cos5(x) = [cos2(x)]2 cos(x) = [1 – sin2(x)]2 cos(x) Now we can substitute u = sin(x), so du = cos(x) dx, and we get ∫cos5(x) dx = ∫[1 – u2]2 du= ∫(1 – 2 u2 + u4) du= u – 2/3 u3 + 1/5 u5 = sin(x) – 2/3 sin3(x) + 1/5 sin5(x)

 

3.Find ∫cos2(x) dx

Answer

1. x/2 + 1/4 sin(2x) + c

Solution

1. Just as we did in the notes with sin2(x), we can use the formula for cos(2x) to help us out: cos(2x) = cos2(x) – sin2(x)= cos2(x) – [1 – cos2(x)] = 2 cos2(x) – 1 Now we turn it around: cos2(x) = 1/2[1 + cos(2x)] This we know how to integrate: ∫cos2(x) dx = ∫1/2[1 + cos(2x)] dx = x/2 + 1/4 sin(2x) + c

 

4. Use the formula ∫(1 – x2)-1/2 dx = arcsin(x) + c to find ∫(4 – x2)-1/2 dx

Answer

1. Arcsin(x/2) + c

Solution

1. Our first goal is to turn the “4” into a “1”, so we pull 4 our of the parentheses: (4 – x2)-1/2= [4 (1 – x2/4)]-1/2= 1/2 [1 – (x/2)2]-1/2 Now we substitute u = x/2, so du = dx / 2, and get ∫(4 – x2)-1/2 dx = ∫[1 – u2]-1/2 du= arcsin(u) + c = arcsin(x/2) + c

 

5. Use trig substitution to find ∫1/(1 + x2) dx

Answer

1. Arctan(x) + c

Solution

1. Use our trig substitution table, and substitute x = tan(u). As written in the notes: 1 + x2 = 1 + tan2(u) = 1/cos2(u) In exercises for Algebra of derivatives we calculated the derivative of tan(x) using the product rule: dx = 1/cos2(u) du The two go very well together: 1/(1 + x2) dx = cos2(u) dx = du Easy to integrate: ∫1/(1 + x2) dx = ∫du = u + c = arctan(x) + c