**Integration** is the reverse process of differentiation. Given the derivative f(x) of a function, integration finds the original function F(x) such that **dF/dx = f(x)**.
**Definition of Antiderivative:** A function F(x) is called an **antiderivative** (or **primitive**) of f(x) on an interval I if **F'(x) = f(x)** for all x ∈ I.
**Key Insight:** Antiderivatives are NOT unique. If F(x) is an antiderivative of f(x), then F(x) + C is also an antiderivative for any constant C, since d/dx[F(x) + C] = F'(x) = f(x).
**Example:** Since d/dx(sin x) = cos x, we have ∫cos x dx = sin x + C. All functions of the form sin x + C are antiderivatives of cos x.
**Important Remark:** If two functions have the same derivative on an interval, they differ by a constant. This justifies why we always add "+ C" (the **constant of integration**) to represent the entire family of antiderivatives.
**Notation:**
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These formulas follow directly from standard derivative formulas. Every student must memorize these:
1. **∫x^n dx = (x^(n+1))/(n+1) + C**, where n ≠ -1
2. **∫1 dx = x + C**
3. **∫cos x dx = sin x + C**
4. **∫sin x dx = -cos x + C**
5. **∫sec² x dx = tan x + C**
6. **∫cosec² x dx = -cot x + C**
7. **∫sec x tan x dx = sec x + C**
8. **∫cosec x cot x dx = -cosec x + C**
9. **∫(1/√(1-x²)) dx = sin⁻¹ x + C**
10. **∫(1/(1+x²)) dx = tan⁻¹ x + C**
11. **∫e^x dx = e^x + C**
12. **∫(1/x) dx = log|x| + C**, where x ≠ 0
13. **∫a^x dx = (a^x)/log a + C**, where a > 0, a ≠ 1
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**Property 1 (Inverse Relationship):**
This shows differentiation and integration undo each other.
**Property 2 (Linearity - Sum Rule):**
**∫[f(x) + g(x)]dx = ∫f(x)dx + ∫g(x)dx**
*Proof:* Both sides give the same derivative by Property 1, so they differ by a constant at most, proving equivalence.
**Example:** ∫(x² + cos x)dx = ∫x² dx + ∫cos x dx = x³/3 + sin x + C
**Property 3 (Linearity - Scalar Multiple):**
**∫k·f(x)dx = k∫f(x)dx**, where k is any real constant
*Proof:* d/dx[k∫f(x)dx] = k·f(x) by Property 1, matching the left side.
**Property 4 (General Linearity):**
**∫[k₁f₁(x) + k₂f₂(x) + ... + kₙfₙ(x)]dx = k₁∫f₁(x)dx + k₂∫f₂(x)dx + ... + kₙ∫fₙ(x)dx**
This combines Properties 2 and 3 for any finite combination of functions and constants.
*Exam Tip:* Always split complicated integrands using linearity before applying standard formulas.
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In this method, we **guess** a function whose derivative is the given integrand.
**Technique:** Look for a function F such that F'(x) = f(x). Common patterns to recognize:
**Example 1:** Find ∫(3x² + 4x³)dx
Recognize that d/dx(x³ + x⁴) = 3x² + 4x³
Therefore, ∫(3x² + 4x³)dx = x³ + x⁴ + C
**Example 2:** Find ∫sin 2x dx
We know d/dx(sin 2x) = 2cos 2x, so d/dx((sin 2x)/2) = cos 2x
But we want the antiderivative of cos 2x to be sin 2x/2... work backwards:
Since d/dx(cos 2x) = -2sin 2x, we have ∫sin 2x dx = -(cos 2x)/2 + C
**Example 3:** Find ∫(1/x) dx for x ≠ 0
Recall: d/dx(log x) = 1/x for x > 0 and d/dx(log(-x)) = 1/x for x < 0
Combined: d/dx(log|x|) = 1/x for all x ≠ 0
Therefore, ∫(1/x)dx = log|x| + C
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When given a specific condition like F(a) = b, use it to find the constant C uniquely.
**Procedure:**
1. Find the general antiderivative: F(x) = (expression in x) + C
2. Substitute the given point: F(a) = (expression at a) + C = b
3. Solve for C
4. Write the specific antiderivative
**Example:** Find F(x) such that F'(x) = 4x³ - 6 and F(0) = 3
General antiderivative: F(x) = x⁴ - 6x + C
Using F(0) = 3: 0 - 0 + C = 3, so C = 3
**Answer: F(x) = x⁴ - 6x + 3**
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When the integrand contains a composite function, **substitution** simplifies it into standard form.
**General Procedure:**
1. Identify u = g(x) such that du = g'(x)dx appears in the integrand
2. Substitute: ∫f(x)dx = ∫f(g(u))·g'(u)du
3. Integrate with respect to u
4. Substitute u back in terms of x
**Key Strategy:** Choose u to be the "inside" function of a composition or the argument of a function whose derivative also appears.
**Example 1:** ∫2x sin(x² + 1)dx
Let u = x² + 1, so du = 2x dx
∫2x sin(x² + 1)dx = ∫sin u du = -cos u + C = **-cos(x² + 1) + C**
**Example 2:** ∫sin mx dx
Let u = mx, so du = m dx, giving dx = du/m
∫sin mx dx = ∫sin u · (du/m) = (1/m)∫sin u du = -(cos u)/m + C = **-(cos mx)/m + C**
**Example 3:** ∫(2x)/(x² + 1) dx
Let u = x² + 1, so du = 2x dx
∫(2x)/(x² + 1) dx = ∫(du/u) = log|u| + C = **log(x² + 1) + C** (since x² + 1 > 0)
**Common Mistakes:**
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These frequently appear in exams and are derived using substitution:
**∫tan x dx = log|sec x| + C**
*Derivation:* ∫(sin x)/(cos x) dx. Let u = cos x, du = -sin x dx.
∫(sin x)/(cos x) dx = -∫(du/u) = -log|u| + C = -log|cos x| + C = log|sec x| + C ✓
**∫cot x dx = log|sin x| + C**
*Derivation:* ∫(cos x)/(sin x) dx. Let u = sin x, du = cos x dx.
∫(cos x)/(sin x) dx = ∫(du/u) = log|u| + C = **log|sin x| + C**
**∫sec x dx = log|sec x + tan x| + C**
*Derivation:* Multiply by (sec x + tan x)/(sec x + tan x):
∫sec x · (sec x + tan x)/(sec x + tan x) dx. Let u = sec x + tan x, du = sec x(tan x + sec x)dx.
∫sec(sec x + tan x)/(sec x + tan x) dx = ∫du/u = **log|sec x + tan x| + C**
**∫cosec x dx = log|cosec x - cot x| + C** or **-log|cosec x + cot x| + C**
*Derivation:* Similar to sec x, using (cosec x - cot x)/(cosec x - cot x):
∫cosec x · (cosec x - cot x)/(cosec x - cot x) dx. Let u = cosec x + cot x, du = -cosec x(cosec x + cot x)dx.
This gives ∫(-du/u) = **log|cosec x - cot x| + C**
**Exam Note:** These four results should be memorized as they appear frequently and don't follow obviously from standard formulas.
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When the integrand is a **rational function** (ratio of polynomials), **partial fractions** decompose it into simpler fractions that integrate easily.
**Form:** If denominator = (x - a)(x - b)(x - c) with a, b, c distinct,
**Decompose:** P(x)/[(x-a)(x-b)(x-c)] = **A/(x-a) + B/(x-b) + C/(x-c)**
**Method to find A, B, C:**
1. Multiply both sides by the denominator: P(x) = A(x-b)(x-c) + B(x-a)(x-c) + C(x-a)(x-b)
2. Substitute x = a: P(a) = A(a-b)(a-c) → find A
3. Substitute x = b: P(b) = B(b-a)(b-c) → find B
4. Substitute x = c: P(c) = C(c-a)(c-b) → find C
**Example:** ∫(1)/[(x-1)(x-2)]dx
Decompose: 1/[(x-1)(x-2)] = A/(x-1) + B/(x-2)
Multiply by denominator: 1 = A(x-2) + B(x-1)
∫(1)/[(x-1)(x-2)]dx = ∫[-1/(x-1) + 1/(x-2)]dx = -log|x-1| + log|x-2| + C = **log|(x-2)/(x-1)| + C**
**Form:** If denominator has (x - a)^n,
**Decompose:** P(x)/[(x-a)^n] = **A₁/(x-a) + A₂/(x-a)² + ... + Aₙ/(x-a)^n** + (terms from other factors)
**Example:** ∫1/[(x-1)²]dx
1/(x-1)² = A/(x-1) + B/(x-1)²
Multiply: 1 = A(x-1) + B
∫1/(x-1)² dx = ∫1/(x-1)² dx = **-1/(x-1) + C**
**Form:** If denominator has (ax² + bx + c) that doesn't factor over reals (discriminant < 0),
**Decompose:** (Ex + F)/(ax² + bx + c), then use substitution to integrate
**Example:** ∫1/(x² + 1)dx = tan⁻¹(x) + C (this is a standard form)
More generally: ∫(2x + 3)/(x² + 2x + 5)dx
Write: (2x + 3)/(x² + 2x + 5) = (2x + 2)/(x² + 2x + 5) + 1/(x² + 2x + 5)
First integral: ∫(2x + 2)/(x² + 2x + 5)dx = log(x² + 2x + 5) + C₁ (by substitution u = x² + 2x + 5)
Second integral: ∫1/(x² + 2x + 5)dx = ∫1/[(x+1)² + 4]dx. Let u = x+1, so ∫1/(u² + 4)du = (1/2)tan⁻¹(u/2) + C₂ = (1/2)tan⁻¹((x+1)/2) + C₂
**Final Answer:** log(x² + 2x + 5) + (1/2)tan⁻¹((x+1)/2) + C
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**Formula:** **∫u dv = uv - ∫v du**
This is derived from the product rule: d/dx(uv) = u(dv/dx) + v(du/dx)
**When to Use:**
**ILATE Rule (Choosing u):** Choose u in order of priority:
1. **I**nverse trig (sin⁻¹, cos⁻¹, tan⁻¹)
2. **L**ogarithmic (log, ln)
3. **A**lgebraic (polynomial, x)
4. **T**rigonometric (sin, cos, tan)
5. **E**xponential (e^x, a^x)
The function appearing first in ILATE should be chosen as u (the part to differentiate).
**Step-by-Step Procedure:**
1. Identify u and dv from the integrand
2. Calculate du (differentiate u) and v (integrate dv)
3. Apply formula: ∫u dv = uv - ∫v du
4. Evaluate the remaining integral
**Example 1:** ∫x e^x dx
Choose u = x (algebraic), dv = e^x dx
Then: du = dx, v = e^x
∫x e^x dx = x·e^x - ∫e^x dx = x e^x - e^x + C = **e^x(x - 1) + C**
**Example 2:** ∫x sin x dx
Choose u = x (algebraic), dv = sin x dx
Then: du = dx, v = -cos x
∫x sin x dx = x(-cos x) - ∫(-cos x)dx = -x cos x + ∫cos x dx = **-x cos x + sin x + C**
**Example 3:** ∫log x dx
Choose u = log x (logarithmic), dv = dx
Then: du = (1/x)dx, v = x
∫log x dx = x log x - ∫x · (1/x)dx = x log x - ∫1 dx = **x log x - x + C** or **x(log x - 1) + C**
**Example 4 (Repeated Application):** ∫x² e^x dx
First: u = x², dv = e^x dx → du = 2x dx, v = e^x
∫x² e^x dx = x² e^x - ∫2x e^x dx
For remaining integral ∫2x e^x dx:
u = 2x, dv = e^x dx → du = 2dx, v = e^x
∫2x e^x dx = 2x e^x - ∫2 e^x dx = 2x e^x - 2e^x
Therefore: ∫x² e^x dx = x² e^x - (2x e^x - 2e^x) + C = **e^x(x² - 2x + 2) + C**
**Common Pattern:** ∫P(x) e^x dx where P is polynomial of degree n requires n applications of integration by parts.
**Special Cyclic Case:** Some integrals like ∫e^x sin x dx require the formula to be applied twice, then solve for the integral:
∫e^x sin x dx = e^x sin x - ∫e^x cos x dx (first integration by parts)
∫e^x cos x dx = e^x cos x - ∫(-e^x sin x)dx = e^x cos x + ∫e^x sin x dx (second integration by parts)
Substituting back:
∫e^x sin x dx = e^x sin x - (e^x cos x + ∫e^x sin x dx)
2∫e^x sin x dx = e^x(sin x - cos x)
**∫e^x sin x dx = [e^x(sin x - cos x)]/2 + C**
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**Standard Form:** **∫1/√(a² - x²) dx = sin⁻¹(x/a) + C**, where |x| < a
**More Generally:**
*Derivation of second formula:*
Let x = a sin θ, dx = a cos θ dθ, √(a² - x²) = a cos θ
∫√(a² - x²) dx = ∫a cos θ · a cos θ dθ = a²∫cos² θ dθ = a²∫[(1 + cos 2θ)/2]dθ
= (a²/2)[θ + (sin 2θ)/2] + C = (a²/2)θ + (a²/4)(2 sin θ cos θ) + C
= (a²/2)sin⁻¹(x/a) + (a²/4) · (2x/a) · √(a² - x²)/a + C
= (a²/2)sin⁻¹(x/a) + (x/2)√(a² - x²) + C ✓
**Example:** ∫√(9 - x²) dx with a = 3
= (x/2)√(9 - x²) + (9/2)sin⁻¹(x/3) + C
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**For √(x² + a²):**
**∫1/√(x² + a²) dx = log|x + √(x² + a²)| + C** or **sinh⁻¹(x/a) + C**
**∫√(x² + a²) dx = (x/2)√(x² + a²) + (a²/2)log|x + √(x² + a²)| + C**
*Derivation:* Let x = a tan θ, dx = a sec² θ dθ, √(x² + a²) = a sec θ
∫1/√(x² + a²) dx = ∫(a sec² θ)/(a sec θ) dθ = ∫sec θ dθ = log|sec θ + tan θ| + C
Since tan θ = x/a and sec θ = √(x² + a²)/a:
= log|√(x² + a²)/a + x/a| + C = log|x + √(x² + a²)| - log a + C
The -log a merges into C, giving **log|x + √(x² + a²)| + C**
**For √(x² - a²):**
**∫1/√(x² - a²) dx = log|x + √(x² - a²)| + C** (for |x| > a)
**∫√(x² - a²) dx = (x/2)√(x² - a²) - (a²/2)log|x + √(x² - a²)| + C**
**Example:** ∫√(x² - 4) dx with a = 2
= (x/2)√(x² - 4) - (4/2)log|x + √(x² - 4)| + C
= (x/2)√(x² - 4) - 2 log|x + √(x² - 4)| + C
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**Definition:** The **definite integral** of f from a to b is:
**∫ₐᵇ f(x)dx = F(b) - F(a) = [F(x)]ₐᵇ**
where F is any antiderivative of f, and a is the **lower limit** while b is the **upper limit**.
**Geometric Meaning:** Represents the **net area** between the curve y = f(x) and the x-axis from x = a to x = b. Area above the x-axis is positive; area below is negative.
**Important:** Unlike indefinite integrals, definite integrals have NO constant of integration C, because the constants cancel:
∫ₐᵇ f(x)dx = [F(x) + C]ₐᵇ = [F(b) + C] - [F(a) + C] = F(b) - F(a)
**Example:** ∫₀^π sin x dx = [-cos x]₀^π = (-cos π) - (-cos 0) = -(-1) - (-1) = 1 + 1 = **2**
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**Property 1:** **∫ₐᵃ f(x)dx = 0**
When lower and upper limits are equal, the integral is zero.
**Property 2:** **∫ₐᵇ f(x)dx = -∫ᵇₐ f(x)dx**
Swapping limits changes the sign.
**Property 3:** **∫ₐᵇ f(x)dx = ∫ₐᶜ f(x)dx + ∫ᶜᵇ f(x)dx**, where a < c < b
The integral over [a,b] equals the sum of integrals over subintervals [a,c] and [c,b]. This is useful when integrand changes form at c.
**Property 4:** **∫ₐᵇ k·f(x)dx = k∫ₐᵇ f(x)dx** (scalar multiple)
**Property 5:** **∫ₐᵇ [f(x) + g(x)]dx = ∫ₐᵇ f(x)dx + ∫ₐᵇ g(x)dx** (sum rule)
**Property 6:** **∫ₐᵇ f(x)dx = ∫ₐᵇ f(a+b-x)dx** (important property - proof below)
*Proof:* Let u = a + b - x, then du = -dx. When x = a, u = b; when x = b, u = a.
∫ₐᵇ f(a+b-x)dx = ∫ᵇₐ f(u)(-du) = ∫ₐᵇ f(u)du = ∫ₐᵇ f(x)dx ✓
*Use Case:* This property is powerful for simplifying trigonometric integrals:
∫₀^π x sin x dx: Use property with a = 0, b = π:
∫₀^π x sin x dx = ∫₀^π (π - x)sin x dx
Adding both:
2∫₀^π x sin x dx = ∫₀^π [x sin x + (π - x)sin x]dx = π∫₀^π sin x dx = π[-cos x]₀^π = π · 2 = 2π
Therefore: ∫₀^π x sin x dx = **π**
**Property 7:** If f is even [f(-x) = f(x)]: **∫₋ₐᵃ f(x)dx = 2∫₀ᵃ f(x)dx**
If f is odd [f(-x) = -f(x)]: **∫₋ₐᵃ f(x)dx = 0**
*Application:* ∫₋₁¹ x³ dx = 0 (odd function), ∫₋₁¹ x² dx = 2∫₀¹ x² dx = 2 · [x³/3]₀¹ = 2/3 (even function)
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Sometimes integrals cannot be evaluated using antiderivatives. We use the **definition by limits**:
**∫ₐᵇ f(x)dx = lim_{n→∞} Σ(i=1 to n) f(xᵢ)Δx**
where Δx = (b - a)/n and xᵢ = a + i·Δx
**Step-by-Step Approach:**
1. Divide [a,b] into n equal parts: Δx = (b - a)/n
2. Choose right endpoints: xᵢ = a + i·Δx for i = 1, 2, ..., n
3. Form Riemann sum: S = Σ(i=1 to n) f(a + i·Δx)·Δx
4. Simplify the sum using formulas (see below)
5. Take limit as n → ∞
**Standard Sum Formulas:**
**Example:** Evaluate ∫₀¹ x² dx by limit of sum
Δx = (1 - 0)/n = 1/n, xᵢ = i/n
S_n = Σ(i=1 to n) (i/n)² · (1/n) = (1/n³)Σ(i=1 to n) i²
= (1/n³) · n(n+1)(2n+1)/6 = (n+1)(2n+1)/(6n²)
lim_{n→∞} (n+1)(2n+1)/(6n²) = lim_{n→∞} (2n² + 3n + 1)/(6n²) = 2/6 = **1/3**
This matches ∫₀¹ x² dx = [x³/3]₀¹ = 1/3 ✓
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The **area bounded** by the curve y = f(x), the x-axis, and the lines x = a and x = b (where f(x) ≥ 0) is:
**A = ∫ₐᵇ f(x)dx**
If f(x) < 0 on some portion, we take absolute value of the integral over that portion:
**A = ∫ₐᶜ |f(x)|dx + ∫ᶜᵇ |f(x)|dx** where c is the zero of f in [a,b]
**Example:** Find the area under y = x² from x = 0 to x = 2
A = ∫₀² x² dx = [x³/3]₀² = 8/3 - 0 = **8/3 square units**
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If two curves y = f(x) and y = g(x) intersect at x = a and x = b with **f(x) ≥ g(x)** on [a,b], the area between them is:
**A = ∫ₐᵇ [f(x) - g(x)]dx**
**Procedure:**
1. Find intersection points by solving f(x) = g(x)
2. Determine which curve is above on each interval
3. Integrate the difference (upper - lower)
**Example:** Find the area between y = x² and y = x from x = 0 to x = 1
Find intersections: x² = x → x(x-1) = 0 → x = 0 or x = 1
Check which is above: at x = 0.5, line gives 0.5, parabola gives 0.25, so y = x is above
A = ∫₀¹ (x - x²)dx = [x²/2 - x³/3]₀¹ = (1/2 - 1/3) - 0 = **1/6 square units**
**For Parabola and Line:** When integrating with respect to x may be difficult, switch to **integration with respect to y**:
A = ∫ᶜᵈ [g(y) - f(y)]dy where the curves are expressed as x = f(y) and x = g(y)
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**Standard Integrals:**
**Integration Methods:**
Q1. If the derivative of F(x) is f(x), which statement defines the indefinite integral ∫f(x)dx?
Answer: A — By definition, the indefinite integral gives the family of all antiderivatives, differing by arbitrary constant C.
Q2. Evaluate ∫(5x⁴ + 3x² - 2)dx.
Answer: A — Apply power rule term-by-term: ∫5x⁴dx = 5·x⁵/5 = x⁵; ∫3x²dx = 3·x³/3 = x³; ∫-2dx = -2x; add C.
Q3. Which of the following is correct regarding antiderivatives of the same function?
Answer: B — If F and G both satisfy F'(x) = G'(x) = f(x), then (F-G)' = 0, so F(x) - G(x) = constant.
Q4. What is ∫(1/x)dx for x > 0?
Answer: C — The antiderivative of 1/x is log|x| + C; absolute value is used because log is defined for negative x too.
Q5. If F(x) = x³ + 2x + 5, what is ∫f(x)dx where f(x) = F'(x)?
Answer: B — F'(x) = 3x² + 2, so ∫(3x² + 2)dx = x³ + 2x + C; the original function F plus constant C.
Q6. Which property correctly states the relationship between differentiation and integration?
Answer: B — Differentiation of an indefinite integral recovers the original integrand f(x); the constant C differentiates to zero.
Q7. When evaluating ∫x⁻¹dx, why cannot we use the power rule ∫xⁿdx = xⁿ⁺¹/(n+1) + C?
Answer: B — The power rule fails when n = -1 since xⁿ⁺¹/(n+1) = x⁰/0, which is undefined; instead, ∫x⁻¹dx = log|x| + C.
Q8. Consider two statements: (I) Every continuous function has an antiderivative. (II) The constant of integration C represents infinitely many antiderivatives. Which is correct?
Answer: A — Both statements are true: (I) is a theorem in calculus; (II) follows because differentiating constant gives zero, allowing C to vary.
Q9. If ∫f(x)dx = x⁴/4 - x²/2 + 7, what is f(x)?
Answer: A — Differentiate: d/dx[x⁴/4 - x²/2 + 7] = 4x³/4 - 2x/2 + 0 = x³ - x; constant 7 differentiates to zero.
Q10. A common mistake: Student A writes ∫sin(x)dx = cos(x) + C. Is this correct? If not, what is the error?
Answer: B — Since d/dx[-cos(x)] = sin(x), the antiderivative of sin(x) is -cos(x) + C, not cos(x) + C; neglecting the negative is a critical exam error.
What is an antiderivative (primitive) of a function?
A function F(x) such that F'(x) = f(x), i.e., when differentiated it gives the original function f(x).
Why does the constant of integration C appear in indefinite integrals?
Because the derivative of any constant is zero, so infinitely many functions differing only by a constant have the same derivative.
State the Power Rule for Integration: ∫xⁿdx = ?
∫xⁿdx = xⁿ⁺¹/(n+1) + C for n ≠ -1; when n = -1, use ∫(1/x)dx = log|x| + C.
What is ∫cos(x)dx and ∫sin(x)dx?
∫cos(x)dx = sin(x) + C and ∫sin(x)dx = -cos(x) + C.
State the fundamental property: d/dx[∫f(x)dx] = ?
d/dx[∫f(x)dx] = f(x), showing differentiation and integration are inverse processes.
Find ∫eˣdx.
∫eˣdx = eˣ + C.
What is ∫(1/x)dx for x ≠ 0?
∫(1/x)dx = log|x| + C; absolute value is essential because log is undefined for negative arguments.
Evaluate ∫(3x² + 2x + 5)dx.
∫(3x² + 2x + 5)dx = 3·x³/3 + 2·x²/2 + 5x + C = x³ + x² + 5x + C.
If F and G both have the same derivative on interval I, how do they relate?
F(x) - G(x) = C (constant), meaning any two antiderivatives of the same function differ by only a constant.
Why do we write ∫f(x)dx = F(x) + C instead of just F(x)?
The '+ C' represents the entire family of antiderivatives; omitting it means missing infinitely many valid solutions.
Define antiderivative and indefinite integral. If F'(x) = f(x), write the relationship between F and the indefinite integral ∫f(x)dx. [2 marks]
State that antiderivative is function whose derivative gives f(x); write ∫f(x)dx = F(x) + C showing why arbitrary constant C appears (derivative of constant is zero).
Prove that if two functions F(x) and G(x) have the same derivative on an interval I, then F(x) - G(x) = C, where C is a constant. Use this result to justify why the indefinite integral includes an arbitrary constant. [5 marks]
Define h(x) = F(x) - G(x); show h'(x) = F'(x) - G'(x) = 0 on I, implying h is constant; conclude that any two antiderivatives differ by constant only, so indefinite integral is entire family F(x) + C.
Using standard integration formulas and the concept of antiderivatives, evaluate ∫(6x⁵ + 4cos(x) - 3/x + eˣ)dx. Show all working steps and verify your answer by differentiation. [6 marks]
Apply power rule to 6x⁵, use ∫cos(x)dx = sin(x) + C, use ∫(1/x)dx = log|x| + C, and ∫eˣdx = eˣ + C; combine to get result, then differentiate to confirm it equals the original integrand.
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