Integration

Integration

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Integration is the reverse of differentiation. If d/dx[F(x)] = f(x), then F(x) is an antiderivative of f(x) and we write ∫f(x)dx = F(x) + C, where C is the constant of integration. The definite integral ∫ₐᵇ f(x)dx = F(b) − F(a) by the Fundamental Theorem of Calculus (FTC). Must-know primitives:

• ∫xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ −1)
• ∫(1/x) dx = log|x| + C
• ∫eˣ dx = eˣ + C
• ∫sin x dx = −cos x + C ; ∫cos x dx = sin x + C
• ∫sec²x dx = tan x + C ; ∫cosec²x dx = −cot x + C

CUET MCQ traps: omitting +C, treating 1/x like xⁿ (giving log|x|, not 0), and sign slip on ∫sin x. Always remember: integrand dx = antiderivative evaluated, limits substituted.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Concept and Notation

The indefinite integral ∫f(x)dx returns a family of functions differing by a constant — geometrically, a vertical family of antiderivative curves. The integrand is f(x), the variable of integration is x, and dx indicates the variable being summed in infinitesimal pieces. A definite integral ∫ₐᵇ f(x)dx is a number: the signed area between the curve and the x-axis from x = a to x = b. Limits may be swapped: ∫ₐᵇ = −∫ᵦᵃ, and the interval may be split: ∫ₐᵇ = ∫ₐᶜ + ∫ᶜᵇ.

Fundamental Theorem of Calculus (FTC)

If F is any antiderivative of f on [a, b], then ∫ₐᵇ f(x)dx = F(b) − F(a). The FTC also yields ∫ₐˣ f(t)dt = f(x), linking differentiation and integration as inverse operations. CUET frequently tests FTC as a one-step evaluation after recognising the antiderivative.

Methods of Evaluation

Method	When to use	Key step
Substitution (u = g(x))	Composite function f(g(x))·g′(x)	du = g′(x)dx; change limits if definite
Integration by parts	Product of unlike types (x·sin x, log x, eˣ·x)	∫u dv = uv − ∫v du; use ILATE for u-priority
Partial fractions	Rational P(x)/Q(x), deg P < deg Q	Factor Q, split into A/(x−a) + B/(x−b) + …
Trig identities	∫sin²x, ∫cos²x, ∫sec x·tan x	Half-angle, double-angle conversions

Standard Forms (must memorise)

Inverse trig: ∫dx/√(a²−x²) = sin⁻¹(x/a) + C ; ∫dx/(a²+x²) = (1/a)tan⁻¹(x/a) + C ; ∫dx/(x√(x²−a²)) = (1/a)sec⁻¹|x/a| + C. Also ∫aˣ dx = aˣ/log a + C and ∫dx/(1+x²) = tan⁻¹x + C.

Exam Patterns in CUET

Expect one-mark MCQs evaluating ∫₀^{π/2} sin x dx, computing area using ∫ₐᵇ y dx, or applying substitution like ∫ 2x/(1+x²) dx. Assertion-reason items often distinguish between indefinite (+C) and definite (no C) outcomes.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Substitution — the Workhorse

The substitution rule states ∫ f(g(x))·g′(x) dx = ∫ f(u) du with u = g(x). For definite integrals, change the limits too: when x = a, u = g(a); when x = b, u = g(b). A classic trap is differentiating inside the integral (treating dy/dx as a cancelable fraction). It is not a fraction — substitution is a change of variable, not algebraic division. Example: ∫ 2x cos(x²) dx. Let u = x², du = 2x dx ⇒ ∫ cos u du = sin u + C = sin(x²) + C.

Integration by Parts — ILATE in Action

ILATE orders u: Inverse trig → Log → Algebraic → Trig → Exponential. For ∫ x eˣ dx, take u = x, dv = eˣ dx ⇒ du = dx, v = eˣ, so the integral = x eˣ − ∫ eˣ dx = eˣ(x − 1) + C. CUET sometimes offers a “definite by parts” item like ∫₀¹ x² log(1+x) dx — set u = log(1+x), dv = x² dx.

Definite Integral Properties and Area

For a continuous f on [a, b]:

• ∫ₐᵃ f(x) dx = 0
• ∫ₐᵇ f(x) dx = −∫ᵦᵃ f(x) dx
• ∫ₐᵇ [f(x) ± g(x)] dx = ∫ₐᵇ f dx ± ∫ₐᵇ g dx
• ∫₀^{2a} f(x) dx = ∫₀ᵃ [f(x) + f(2a − x)] dx (King’s / symmetry property)
• Area bounded by y = f(x) ≥ 0 and the x-axis from a to b equals ∫ₐᵇ f(x) dx

Improper Integrals and Special Forms

When one limit is ∞, write ∫ₐ^∞ f(x) dx = lim_{t→∞} ∫ₐᵗ f(x) dx if the limit exists. ∫₀^∞ e⁻ˣ dx = 1 (used in probability). Beta-function style integrals ∫₀^π/2 sinⁿx dx and ∫₀^π/2 cosⁿx dx follow Wallis’ reduction and are occasionally tested in CUET domain knowledge items.

Common Mistakes

1. Writing ∫(1/x) dx = x⁰/0 — undefined; correct form is log|x| + C.
2. Dropping +C in an indefinite answer.
3. Sign error: ∫sin x dx = −cos x + C (not +cos x).
4. For ∂u substitution in definite integrals: forgetting to convert limits.
5. Applying parts twice and getting tangled signs — keep du and v aligned.

Worked Example

Evaluate ∫₀² x √(x+1) dx. Let t = x+1 ⇒ x = t−1, dx = dt; limits 1 → 3. Integral becomes ∫₁³ (t−1)√t dt = ∫₁³ (t^{3/2} − t^{1/2}) dt = [⅖ t^{5/2} − ⅔ t^{3/2}]₁³ = (⅖·9√3 − ⅔·3√3) − (⅖ − ⅔) = (18√3/5 − 2√3) + (1/6) = (18√3 − 10√3)/5 + 1/6 = 8√3/5 + 1/6.

Practice Prompts

1. Compute ∫ (3x² + 2)/(x³ + 2x + 5) dx using u-substitution.
2. Evaluate ∫₀^{π/4} (sec²x · log(tan x)) dx using integration by parts.

Connection to Other CUET Topics

Integration links directly to Differential Equations (separable equations solve by ∫dx/x = ∫dy/y), Application of Integrals (area between curves, volume by disk method), and Probability (continuous distributions integrate density to 1). Mastery here boosts three separate CUET chapters.

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