Author Archives: mh225

Numerical Sequences and Series: The comparison test

Notes PDF
More Challenging Problems 

 

Determine if the series, a1 + a2 + a3 + a4 + … converges or diverges, where:

1. an= 1/(3n + 2)

Answer

1. Converges

Solution

1. First, observe that 0 < 1/(3n + 2) < 1/3n The series with bn = 1/3n is a geometric series with r= 1/3 and so converges. Then the series with an= 1/(3n + 2) converges by the comparison test.

 

2. an= 1/(n – 1)

Answer

1. Diverges

Solution

1. First, observe that 1/(n – 1) > 1/n The series with bn= 1/n is the harmonic series and so diverges. Then the series with an = 1/(n – 1) diverges by the comparison test.

 

3. an = 1/(n2 + n)

Answer

1. Converges

Solution

1. This is a little trickier, because we could compare this an with 1/n (the harmonic series, so diverges) or with 1/n2 (a p-series with p = 2 > 1, so converges). Which should be used?
Note: 1/(n2 + n) < 1/n and 1/(n2 + n) < 1/n2 Being less than a diverging series tells us nothing, but being less than a converging series tells us that the series of an converges.

 

4. an = 1/(3n – 2)

Answer

1. Converges

Solution

1. The Comparison Test cannot be applied, because 1/(3n – 2) > 1/3n and although the geometric series ∑ 1/3n converges, being greater than a converging series tells us nothing. We can use the Limit Comparison Test, because as n → ∞, (1/(3n – 2)/(1/3n) = (3n)/(3n – 2) = 1/(1 – 2/3n) → 1 Then by the Limit Comparison Test, ∑ 1/(3n – 2) converges because ∑ 1/3n converges.

 

5. an = (√(n3 + 4))/(n2 + 2n + 3)

Answer

1. Diverges

Solution

1. The first issue is to find what series to use for comparison. For large n the numerator √(n3 + 4) is close to n3/2. For large n the denominator n2 + 2n + 3 is close to n2. Then the fraction an is close to n3/2/n2 = 1/n1/2. This is the general term of the series we use for comparison. As n → ∞, (√(n3 + 4))/(n2 + 2n + 3) / (1/n1/2) = (n3/2 √(1 + 4/n3)) / (n2(1 + 2/n + 3/n2)) ⋅ n1/2 = (√(1 + 4/n3)) / (1 + 2/n + 3/n2) → 1 Being a p-series with p = 1/2 < 1, ∑1/n1/2 diverges, so the original series diverges by the Limit Comparison Test.

 

6. an = (3n + 2)/(5n + 4)

Answer

1. Converges

Solution

1. For large n the ratio (3n + 2)/(5n + 4) is very close to 3n/5n, the general term of a geometric series with ratio 3/5, hence converging. As n → ∞, ((3n + 2)/(5n + 4))/(3n/5n) = ((3n(1 + 2/3n))/(5n(1 + 4/5n))⋅(5n/3n) = (1 + 2/3n)/(1 + 4/5n) → 1 Then by the Limit Comparison Test, the original series converges because the geometric series converges.

 

 

Numerical Sequences and Series: The integral test

Notes PDF
More Challenging Problems

Determine if the series, a1 + a2 + a3 + a4 +… converges or diverges, when:

1. an = n/(n2 + 1)

Answer

1. Diverges

Solution

1. f(x)= x/(x2 + 1) is positive, continuous, and decreasing because f'(x)= (1 – x2)/(1 + x2)2 is negative for x > 1. Substituting u= x2 + 1 so du= 2x dx and x dx = (1/2)du, we see ∫x/(x2 + 1) dx = (1/2) ∫du/u = (1/2)ln(u) = (1/2)ln(x2 + 1) This diverges to ∞ as x → ∞, so the series diverges by the Integral Test.

 

2.an= n2/(n3 + 1)

Answer

1. Diverges

Solution

1. f(x)= x2/(x3 + 1) is positive, continuous. Now f'(x)= x(2 – x3)/(1 + x3)2 is negative for x > 21/3≈ 1.2599, so we can apply the Integral test only to the series starting with n= 2. This is fine: the n = 1 term cannot affect the convergence of the series. Substituting u= x3 + 1, so du = 3x2 dx and x2 dx = (1/3)du, we see ∫x2/(x3 + 1) dx = (1/3) ∫du/u = (1/3) ln(u) = (1/3) ln(x3 + 1) This diverges to ∞ as x → ∞, so the series diverges by the Integral Test.

 

3. an= n/(n2 + 1)2

Answer

1. Converges

Solution

1. f(x)= x/(x2 + 1)2 is positive, continuous, and decreasing because f'(x)= (1 – 3x2)/(1 + x2)3is negative for x > 1. Substituting u = x2 + 1 so du = 2x dx and x dx = (1/2)du, we see: ∫x/(x2 + 1)2 dx = (1/2)∫ du/u2 = -(1/2)(1/u) = -(1/2)(1/(x2 + 1)) This converges, so the series converges by the Integral Test.

 

4. an=1/(n ln(n)), starting with n = 2

Answer

1. Diverges

Solution

1. f(x)= 1/(x ln(x)) is positive, continuous, and decreasing because f'(x)= -(1 + ln(x))/(x2ln2(x)) is negative for > 1. Substituting u= ln(x) so du = (1/x)dx, we see ∫1/(x ln(x)) dx = ∫ du/u = ln(u) = ln(ln(x)) This diverges to ∞ as x → ∞, so the series diverges by the Integral Test.

 

Numerical Sequences and Series: Zeno’s Paradox

Notes PDF
More Challenging Problems

1. Find the sum of the series 1 – 1/2 + 1/4 – 1/8 + …

Answer

1. The series sums to 2/3.

Solution

 

2. Find the sum of the series 3 + 3/2 + 3/4 + 3/8 + …

Answer

1. The series sums to 6.

Solution

1. Factoring out a 3, this series is 3⋅(1 + 1/2 + 1/4 + 1/8 + …) The series in parentheses is a geometric series with ratio r = 1/2, converging because |r| < 1. This series sums to 1/(1 – r) = 1/(1 – (1/2)) = 2, so the original series sums to 3⋅2 = 6

 

3. Find the sum of the series 1/9 + 1/27 + 1/81 + 1/243 + …

Answer

1. The series sums to 1/6.

Solution

1. Factoring out a 1/9 from every term, we see the series is (1/9)⋅(1 + 1/3 + 1/9 + 1/27 + …) The series in parentheses is a geometric series with ratio r = 1/3, converging because |r| < 1. This series sums to 1/(1 – r) = 3/2, so the original series sums to (1/9)⋅(3/2) = 1/6

 

4. Find the sum of the series 1 + 2 + 4 + 8 + 16 + … + 128

Answer

1. The series sums to 255

Solution

1. Recall for all r, 1 + r + r2 + … + rN-1= (1 – rN)/(1 – r) Because 128= 27, we see that 1 + 2 + 4 + … + 128 = (1 – 28)/(1 – 2) = (1 – 256)/(1-2) = -255.

 

5. Generalize problem 4 to 1 + 2 + 4 + … + 2n

Answer

1. The series sums to 2n+1– 1

Solution

1. Recall for all r, 1 + r + r2 + … + rN-1= (1 – rN)/(1 – r) Then 1 + 2 + 4 + … + 2n= (1 – 2n+1)/(1 – 2)= 2n+1– 1.

 

Numerical Sequences and Series

Notes PDF                                   
More Challenging Problems

In problems 1 – 4 determine if the sequences converge or diverge. For those that converge, find the limits.

 

1. an= (n2 + 3)/(2n2 – 5)

Answer

1. Converges to 1/2.

Solutions

2. Factoring n2 from numerator and denominator, we find limn→ ∞ (n2 + 3)/(2n2 – 5) = limn<→ ∞(1 + 3/n2)/(2 – 5/n2) = (1 + 0)/(2 – 0) = 1/2

 

2. an = (n2 + 3)/(2n3 – 5)

Answer

1. Converges to 0.

Solution

1. Factoring n3 from numerator and denominator, we find limn → ∞(n2 + 3)/(2n3 – 5)= limn → ∞(1/n + 3/n3)/(2 – 5/n3) = (0 + 0)/(2 – 0) = 0/2 = 0

 

3. an = (n2 + 3)/(2n – 5)

Answer

1. Diverges

Solution

1. Factoring n from numerator and denominator, we find limn → ∞ (n2 + 3)/(2n – 5)= limn → ∞(n + 3/n)/(2 – 5/n) =limn → ∞ n/2= ∞

 

4. an = √(n+1) – √n

Answer

1. Converges to 0

Solution

1. Multiply by (√(n+1) + √n)/(√(n+1) + √n), obtaining limn → ∞ √(n+1) – √n = limn → ∞ (√(n+1) – √n)(√(n+1) + √n)/(√(n+1) + √n) = limn → ∞ (n + 1 – n)/(√(n+1) + √n) = limn → ∞ 1/(√(n+1) + √n) = 0

 

In problems 5 – 8 determine if the series ∑n=1∞(a)n converge or diverge. For those that converge, find the limits. In problem 6, start the sum at n = 2.

5. an = n2/(2n2 + 1)

Answer

1. Diverges

Solution

1. Because limn → ∞ n2/(2n2 + 1)= limn → ∞ 1/(2 + 1/n2) = 1/(2 + 0) = 1/2 the series diverges by the 9th term test.

 

6. an = 2/(n2 – 1)

Answer

1. Converges to 3/2

Solution

1. Apply the method of partial fractions and conclude: 2/(n2 – 1) = 1/(n – 1) – 1/(n + 1) Starting the series from n = 2, the first few terms are (1/(2-1) – 1/(2+1)) + (1/(3-1) – 1/(3+1)) + (1/(4-1) – 1/(4+1)) + (1/(5-1) – 1/(5+1)) + … = (1 – 1/3) + (1/2 – 1/4) + (1/3 – 1/5) + (1/4 – 1/6) + … = 1 + 1/2 = 3/2 because 1/3 occurs twice, once + and once -, 1/4 occurs twice, once + and once -, and so on. Only the 1 and 1/2 are not cancelled by later terms.

 

7. an = n sin(1/n)

Answer

1. Diverges

Solution

1. Note that n sin(1/n) = sin(1/n)/(1/n). Because 1/n → 0 as n → ∞, and recalling limx → 0 sin(x)/x = 1, we see limn → ∞n sin(1/n) = 1. The series diverges by the nth term test.

 

8. an = arctan(n)

Answer

1. Diverges

Solution

1. As n → ∞, arctan(n) → π/2, so the series diverges by the nth term test.

 

 

 

 

More Challenging Problems: Complex numbers and Euler’s formula

Notes PDF
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

Notes PDF

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

Notes PDF

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

Notes PDF
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.