Advanced Integration By Parts: Functional, Error, And Recurrence Problems

1) Let f(x) be a twice-differentiable function with f(0) = 0, f(1) = 1, and f'(1) = 2. Evaluate ∫₀¹ x f''(x) dx.

1

2

0

-1

We use integration by parts. Let u = x and dv = f''(x) dx. Then du = dx and v = f'(x). Applying the formula: ∫₀¹ x f''(x) dx = [x f'(x)]₀¹ - ∫₀¹ f'(x) dx. Evaluate the first term: (1 * f'(1)) - (0 * f'(0)) = 1 * 2 - 0 = 2. Evaluate the second term: ∫₀¹ f'(x) dx is simply f(x) evaluated from 0 to 1, which is f(1) - f(0) = 1 - 0 = 1. Finally, subtract the second term from the first: 2 - 1 = 1.

Explanation

We use integration by parts. Let u = x and dv = f''(x) dx. Then du = dx and v = f'(x). Applying the formula: ∫₀¹ x f''(x) dx = [x f'(x)]₀¹ - ∫₀¹ f'(x) dx. Evaluate the first term: (1 * f'(1)) - (0 * f'(0)) = 1 * 2 - 0 = 2. Evaluate the second term: ∫₀¹ f'(x) dx is simply f(x) evaluated from 0 to 1, which is f(1) - f(0) = 1 - 0 = 1. Finally, subtract the second term from the first: 2 - 1 = 1.

2) Find ∫ x sin(x) dx

X cos(x) + sin(x) + C

-x cos(x) + sin(x) + C

X cos(x) - sin(x) + C

-x cos(x) - sin(x) + C

Using integration by parts, we set u = x and dv = sin(x) dx. This gives du = dx and v = -cos(x). Applying the formula ∫ u dv = uv - ∫ v du, we get ∫ x sin(x) dx = x * (-cos(x)) - ∫ (-cos(x)) dx = -x cos(x) + ∫ cos(x) dx. The integral of cos(x) is sin(x). Adding the constant of integration gives us -x cos(x) + sin(x) + C.

Explanation

Using integration by parts, we set u = x and dv = sin(x) dx. This gives du = dx and v = -cos(x). Applying the formula ∫ u dv = uv - ∫ v du, we get ∫ x sin(x) dx = x * (-cos(x)) - ∫ (-cos(x)) dx = -x cos(x) + ∫ cos(x) dx. The integral of cos(x) is sin(x). Adding the constant of integration gives us -x cos(x) + sin(x) + C.

3) Evaluate ∫ ln(x) dx

X ln(x) + C

Ln(x) + C

X ln(x) - x + C

X ln(x) + x + C

For this integral, we choose u = ln(x) and dv = dx. This gives du = (1/x) dx and v = x. Using integration by parts: ∫ ln(x) dx = x ln(x) - ∫ x * (1/x) dx = x ln(x) - ∫ 1 dx. The integral of 1 with respect to x is simply x. Therefore, we obtain x ln(x) - x + C.

Explanation

For this integral, we choose u = ln(x) and dv = dx. This gives du = (1/x) dx and v = x. Using integration by parts: ∫ ln(x) dx = x ln(x) - ∫ x * (1/x) dx = x ln(x) - ∫ 1 dx. The integral of 1 with respect to x is simply x. Therefore, we obtain x ln(x) - x + C.

4) When integrating ∫ x² eˣ dx using integration by parts, the best choice for u is:

Eˣ

X²

X

2x

In integration by parts, we want to choose u to be a function that becomes simpler when we differentiate it. If we let u = x², then du = 2x dx, which reduces the power of x. If we continue, the next application would have u = x, giving du = dx. This progressive simplification makes the integration manageable. If we instead chose u = eˣ, differentiating would give du = eˣ dx, which doesn't simplify the problem.

Explanation

In integration by parts, we want to choose u to be a function that becomes simpler when we differentiate it. If we let u = x², then du = 2x dx, which reduces the power of x. If we continue, the next application would have u = x, giving du = dx. This progressive simplification makes the integration manageable. If we instead chose u = eˣ, differentiating would give du = eˣ dx, which doesn't simplify the problem.

5) Consider the following erroneous "proof" that 0 = 1: Let I = ∫ (1/x) dx. Use integration by parts with u = 1/x and dv = dx. Then du = -1/x² dx and v = x. I = (1/x)(x) - ∫ x (-1/x²) dx I = 1 - ∫ (-1/x) dx I = 1 + ∫ (1/x) dx I = 1 + I Subtracting I from both sides yields 0 = 1. What is the specific flaw in this logic?

Integration by parts cannot be applied to 1/x.

The integrals I on both sides differ by a constant of integration.

The derivative of 1/x was calculated incorrectly.

The variable x cannot cancel out in the term (1/x)(x).

The flaw lies in treating the indefinite integral ∫ (1/x) dx as a single, unique function. An indefinite integral represents a family of functions that differ by a constant, i.e., ln|x| + C. When we write I = 1 + I, we are actually stating that ln|x| + C₁ = 1 + (ln|x| + C₂). The ln|x| terms cancel, leaving C₁ = 1 + C₂, or C₁ - C₂ = 1. This simply shows that the constants of integration on each side differ by 1, which is perfectly valid. It does not imply that 0 = 1. The error is equating two different antiderivatives from the same family and treating them as identical algebraic variables.

Explanation

The flaw lies in treating the indefinite integral ∫ (1/x) dx as a single, unique function. An indefinite integral represents a family of functions that differ by a constant, i.e., ln|x| + C. When we write I = 1 + I, we are actually stating that ln|x| + C₁ = 1 + (ln|x| + C₂). The ln|x| terms cancel, leaving C₁ = 1 + C₂, or C₁ - C₂ = 1. This simply shows that the constants of integration on each side differ by 1, which is perfectly valid. It does not imply that 0 = 1. The error is equating two different antiderivatives from the same family and treating them as identical algebraic variables.

6) After the first application of integration by parts to ∫ eˣ cos(x) dx with u = cos(x) and dv = eˣ dx, the result is:

Eˣ cos(x) - ∫ eˣ sin(x) dx

Eˣ cos(x) + ∫ eˣ sin(x) dx

Eˣ sin(x) - ∫ eˣ cos(x) dx

Eˣ sin(x) + ∫ eˣ cos(x) dx

We start with u = cos(x) and dv = eˣ dx. Then du = -sin(x) dx and v = eˣ. Applying the integration by parts formula gives us ∫ eˣ cos(x) dx = cos(x) * eˣ - ∫ eˣ * (-sin(x)) dx = eˣ cos(x) + ∫ eˣ sin(x) dx. The key step is correctly handling the negative sign when substituting du = -sin(x) dx.

Explanation

We start with u = cos(x) and dv = eˣ dx. Then du = -sin(x) dx and v = eˣ. Applying the integration by parts formula gives us ∫ eˣ cos(x) dx = cos(x) * eˣ - ∫ eˣ * (-sin(x)) dx = eˣ cos(x) + ∫ eˣ sin(x) dx. The key step is correctly handling the negative sign when substituting du = -sin(x) dx.

7) If ∫ xⁿ ln(x) dx = (x⁴/4) ln(x) - x⁴/16 + C, find the value of n.

2

3

4

5

We perform integration by parts on ∫ xⁿ ln(x) dx. Let u = ln(x) and dv = xⁿ dx. Then du = (1/x) dx and v = x^{n+1}/(n+1). The first term of the parts formula (uv) is ln(x) * x^{n+1}/(n+1). Looking at the given answer, the first term is (x⁴/4) ln(x). Equating the powers of x: n + 1 = 4, so n = 3. Equating the denominators: n + 1 = 4, which is consistent. Therefore, n = 3.

Explanation

We perform integration by parts on ∫ xⁿ ln(x) dx. Let u = ln(x) and dv = xⁿ dx. Then du = (1/x) dx and v = x^{n+1}/(n+1). The first term of the parts formula (uv) is ln(x) * x^{n+1}/(n+1). Looking at the given answer, the first term is (x⁴/4) ln(x). Equating the powers of x: n + 1 = 4, so n = 3. Equating the denominators: n + 1 = 4, which is consistent. Therefore, n = 3.

8) ∫ x³ ln x dx =

(1/4) x⁴ (ln x - 1/4) + C

(1/4) x⁴ ln x + C

X⁴ (ln x - 1) + C

X³ ln x + C

Let u = ln x, dv = x³ dx → du = (1/x) dx, v = (1/4) x⁴. Then ∫ x³ ln x dx = (1/4) x⁴ ln x - ∫ (1/4) x⁴ · (1/x) dx = (1/4) x⁴ ln x - (1/4) ∫ x³ dx = (1/4) x⁴ ln x - (1/4)(1/4) x⁴ + C = (1/4) x⁴ (ln x - 1/4) + C.

Explanation

Let u = ln x, dv = x³ dx → du = (1/x) dx, v = (1/4) x⁴. Then ∫ x³ ln x dx = (1/4) x⁴ ln x - ∫ (1/4) x⁴ · (1/x) dx = (1/4) x⁴ ln x - (1/4) ∫ x³ dx = (1/4) x⁴ ln x - (1/4)(1/4) x⁴ + C = (1/4) x⁴ (ln x - 1/4) + C.

9) Evaluate ∫ x arcsin(x) dx.

(x²/2) arcsin(x) - (1/4) arcsin(x) + (x/4)√(1-x²) + C

(x²/2) arcsin(x) + (1/4) arcsin(x) - (x/4)√(1-x²) + C

(x²/2) arcsin(x) - (1/2)√(1-x²) + C

(x²/2) arcsin(x) + (1/4)√(1-x²) + C

Let u = arcsin(x) and dv = x dx. Then du = 1/√(1-x²) dx and v = x²/2. Integral = (x²/2) arcsin(x) - (1/2) ∫ x² / √(1-x²) dx. To solve ∫ x² / √(1-x²) dx, use trig substitution x = sin(θ), dx = cos(θ) dθ. ∫ sin²(θ)/cos(θ) * cos(θ) dθ = ∫ sin²(θ) dθ = ∫ (1 - cos(2θ))/2 dθ = θ/2 - sin(2θ)/4. Convert back to x: θ = arcsin(x). sin(2θ) = 2sin(θ)cos(θ) = 2x√(1-x²). So the integral is (1/2)arcsin(x) - (1/2)x√(1-x²). Multiply by the (1/2) from the parts formula: (1/4)arcsin(x) - (1/4)x√(1-x²). Subtract this from the first term: (x²/2) arcsin(x) - [(1/4)arcsin(x) - (1/4)x√(1-x²)] + C = (x²/2) arcsin(x) - (1/4) arcsin(x) + (x/4)√(1-x²) + C.

Explanation

Let u = arcsin(x) and dv = x dx. Then du = 1/√(1-x²) dx and v = x²/2. Integral = (x²/2) arcsin(x) - (1/2) ∫ x² / √(1-x²) dx. To solve ∫ x² / √(1-x²) dx, use trig substitution x = sin(θ), dx = cos(θ) dθ. ∫ sin²(θ)/cos(θ) * cos(θ) dθ = ∫ sin²(θ) dθ = ∫ (1 - cos(2θ))/2 dθ = θ/2 - sin(2θ)/4. Convert back to x: θ = arcsin(x). sin(2θ) = 2sin(θ)cos(θ) = 2x√(1-x²). So the integral is (1/2)arcsin(x) - (1/2)x√(1-x²). Multiply by the (1/2) from the parts formula: (1/4)arcsin(x) - (1/4)x√(1-x²). Subtract this from the first term: (x²/2) arcsin(x) - [(1/4)arcsin(x) - (1/4)x√(1-x²)] + C = (x²/2) arcsin(x) - (1/4) arcsin(x) + (x/4)√(1-x²) + C.

10) When evaluating ∫ x⁴ sin(x) dx using repeated integration by parts, the final result is a sum of terms involving powers of x multiplied by trig functions. Which of the following is the specific term in the final solution that contains x²?

12x² cos(x)

-12x² cos(x)

12x² sin(x)

-12x² sin(x)

When applying integration by parts repeatedly to ∫ x⁴ sin(x) dx, we differentiate the polynomial part and integrate the trigonometric part until the polynomial becomes a constant.

Step 1: u = x⁴, dv = sin(x) dx. Result: -x⁴ cos(x) - ∫ -cos(x)(4x³) dx = -x⁴ cos(x) + ∫ 4x³ cos(x) dx.

Step 2: Apply parts to ∫ 4x³ cos(x) dx. u = 4x³, dv = cos(x) dx. Result: 4x³ sin(x) - ∫ sin(x)(12x²) dx.

Current Total: -x⁴ cos(x) + 4x³ sin(x) - ∫ 12x² sin(x) dx.

Step 3: Apply parts to ∫ 12x² sin(x) dx. u = 12x², dv = sin(x) dx. Result: 12x²(-cos(x)) - ∫ -cos(x)(24x) dx.

Substitute this back into the Total (watch the subtraction sign from Step 2):

Total: -x⁴ cos(x) + 4x³ sin(x) - [ -12x² cos(x) + ... ]

Total: -x⁴ cos(x) + 4x³ sin(x) + 12x² cos(x) ...

The term containing x² is positive 12x² cos(x).

Explanation

When applying integration by parts repeatedly to ∫ x⁴ sin(x) dx, we differentiate the polynomial part and integrate the trigonometric part until the polynomial becomes a constant.

Step 1: u = x⁴, dv = sin(x) dx. Result: -x⁴ cos(x) - ∫ -cos(x)(4x³) dx = -x⁴ cos(x) + ∫ 4x³ cos(x) dx.

Step 2: Apply parts to ∫ 4x³ cos(x) dx. u = 4x³, dv = cos(x) dx. Result: 4x³ sin(x) - ∫ sin(x)(12x²) dx.

Current Total: -x⁴ cos(x) + 4x³ sin(x) - ∫ 12x² sin(x) dx.

Step 3: Apply parts to ∫ 12x² sin(x) dx. u = 12x², dv = sin(x) dx. Result: 12x²(-cos(x)) - ∫ -cos(x)(24x) dx.

Substitute this back into the Total (watch the subtraction sign from Step 2):

Total: -x⁴ cos(x) + 4x³ sin(x) - [ -12x² cos(x) + ... ]

Total: -x⁴ cos(x) + 4x³ sin(x) + 12x² cos(x) ...

The term containing x² is positive 12x² cos(x).

11) Evaluate ∫ x² ln(x) dx

(x³/3)ln(x) - x³/9 + C

(x³/3)ln(x) + x³/9 + C

(x²/2)ln(x) - x²/4 + C

X³ ln(x) - x³ + C

Let u = ln(x) and dv = x² dx. Then du = (1/x) dx and v = x³/3. Applying integration by parts: ∫ x² ln(x) dx = (x³/3)ln(x) - ∫ (x³/3)(1/x) dx = (x³/3)ln(x) - ∫ (x²/3) dx = (x³/3)ln(x) - (⅓)(x³/3) + C = (x³/3)ln(x) - x³/9 + C.

Explanation

Let u = ln(x) and dv = x² dx. Then du = (1/x) dx and v = x³/3. Applying integration by parts: ∫ x² ln(x) dx = (x³/3)ln(x) - ∫ (x³/3)(1/x) dx = (x³/3)ln(x) - ∫ (x²/3) dx = (x³/3)ln(x) - (⅓)(x³/3) + C = (x³/3)ln(x) - x³/9 + C.

12) Evaluate ∫₀^π x² sin(x) dx

π² - 4

π² + 4

π²

2π

First application: u = x², dv = sin(x) dx gives du = 2x dx, v = -cos(x). So ∫₀^π x² sin(x) dx = [-x² cos(x)]₀^π + 2∫₀^π x cos(x) dx. For the remaining integral, second application: u = x, dv = cos(x) dx gives du = dx, v = sin(x). So ∫₀^π x cos(x) dx = [x sin(x)]₀^π - ∫₀^π sin(x) dx = 0 - [-cos(x)]₀^π = -[(-cos(π) + cos(0))] = -[(1 + 1)] = -2. Now evaluate the first term: [-x² cos(x)]₀^π = -π² cos(π) = -π²(-1) = π². Final result: π² + 2(-2) = π² - 4.

Explanation

First application: u = x², dv = sin(x) dx gives du = 2x dx, v = -cos(x). So ∫₀^π x² sin(x) dx = [-x² cos(x)]₀^π + 2∫₀^π x cos(x) dx. For the remaining integral, second application: u = x, dv = cos(x) dx gives du = dx, v = sin(x). So ∫₀^π x cos(x) dx = [x sin(x)]₀^π - ∫₀^π sin(x) dx = 0 - [-cos(x)]₀^π = -[(-cos(π) + cos(0))] = -[(1 + 1)] = -2. Now evaluate the first term: [-x² cos(x)]₀^π = -π² cos(π) = -π²(-1) = π². Final result: π² + 2(-2) = π² - 4.

13) Evaluate ∫ sec³(x) dx

(sec(x)tan(x) + ln|sec(x) + tan(x)|)/2 + C

Sec(x)tan(x) + C

(sec(x)tan(x) - ln|sec(x) + tan(x)|)/2 + C

Ln|sec(x) + tan(x)| + C

Let I = ∫ sec³(x) dx. Use integration by parts with u = sec(x) and dv = sec²(x) dx. Then du = sec(x)tan(x) dx and v = tan(x). So I = sec(x)tan(x) - ∫ sec(x)tan²(x) dx. Using the identity tan²(x) = sec²(x) - 1, we get I = sec(x)tan(x) - ∫ sec(x)(sec²(x) - 1) dx = sec(x)tan(x) - ∫ sec³(x) dx + ∫ sec(x) dx = sec(x)tan(x) - I + ln|sec(x) + tan(x)|. Rearranging: 2I = sec(x)tan(x) + ln|sec(x) + tan(x)|. Therefore, I = (sec(x)tan(x) + ln|sec(x) + tan(x)|)/2 + C.

Explanation

Let I = ∫ sec³(x) dx. Use integration by parts with u = sec(x) and dv = sec²(x) dx. Then du = sec(x)tan(x) dx and v = tan(x). So I = sec(x)tan(x) - ∫ sec(x)tan²(x) dx. Using the identity tan²(x) = sec²(x) - 1, we get I = sec(x)tan(x) - ∫ sec(x)(sec²(x) - 1) dx = sec(x)tan(x) - ∫ sec³(x) dx + ∫ sec(x) dx = sec(x)tan(x) - I + ln|sec(x) + tan(x)|. Rearranging: 2I = sec(x)tan(x) + ln|sec(x) + tan(x)|. Therefore, I = (sec(x)tan(x) + ln|sec(x) + tan(x)|)/2 + C.

14) Let Iₙ = ∫ xⁿ eˣ dx for n ≥ 0. Which recurrence relation does Iₙ satisfy?

Iₙ = xⁿ eˣ + n Iₙ₋₁

Iₙ = xⁿ eˣ - n Iₙ₋₁

Iₙ = xⁿ⁻¹ eˣ + n Iₙ₋₁

Iₙ = n xⁿ⁻¹ eˣ - Iₙ₋₁

Using integration by parts on Iₙ = ∫ xⁿ eˣ dx, we set u = xⁿ and dv = eˣ dx. This gives du = n xⁿ⁻¹ dx and v = eˣ. Applying the formula: Iₙ = xⁿ eˣ - ∫ eˣ * n xⁿ⁻¹ dx = xⁿ eˣ - n ∫ xⁿ⁻¹ eˣ dx = xⁿ eˣ - n Iₙ₋₁. This recurrence relation is useful for evaluating integrals with higher powers of x by reducing them step by step to simpler integrals.

Explanation

Using integration by parts on Iₙ = ∫ xⁿ eˣ dx, we set u = xⁿ and dv = eˣ dx. This gives du = n xⁿ⁻¹ dx and v = eˣ. Applying the formula: Iₙ = xⁿ eˣ - ∫ eˣ * n xⁿ⁻¹ dx = xⁿ eˣ - n ∫ xⁿ⁻¹ eˣ dx = xⁿ eˣ - n Iₙ₋₁. This recurrence relation is useful for evaluating integrals with higher powers of x by reducing them step by step to simpler integrals.

15) Evaluate ∫ e^{√x} dx

2e^{√x}(√x - 1) + C

E^{√x} + C

√x e^{√x} + C

E^{√x}/√x + C

Use substitution first: let t = √x, so x = t² and dx = 2t dt. The integral becomes ∫ e^t * 2t dt = 2∫ t e^t dt. Now apply integration by parts to ∫ t e^t dt with u = t and dv = e^t dt, giving du = dt and v = e^t. So ∫ t e^t dt = t e^t - ∫ e^t dt = t e^t - e^t + C = e^t(t - 1) + C. Substituting back: 2e^t(t - 1) + C = 2e^{√x}(√x - 1) + C.

Explanation

Use substitution first: let t = √x, so x = t² and dx = 2t dt. The integral becomes ∫ e^t * 2t dt = 2∫ t e^t dt. Now apply integration by parts to ∫ t e^t dt with u = t and dv = e^t dt, giving du = dt and v = e^t. So ∫ t e^t dt = t e^t - ∫ e^t dt = t e^t - e^t + C = e^t(t - 1) + C. Substituting back: 2e^t(t - 1) + C = 2e^{√x}(√x - 1) + C.

Advanced Integration by Parts: Functional, Error, and Recurrence Problems