8.2 Integration & Applications

Why Integration Matters for PAF CAE Maths

After differentiation, the inverse operation—integration—is a standard FSc Pre-Engineering topic and a frequent MCQ source on CAE-style academic papers. You will not be asked to write long ε–δ proofs of the Fundamental Theorem; you will be asked to find antiderivatives, evaluate definite integrals quickly, and compute areas under or between curves.

Antiderivatives and the Indefinite Integral

A function F is an antiderivative of f on an interval if F′(x) = f(x) for every x in that interval. The indefinite integral collects all antiderivatives:

[\int f(x), dx = F(x) + C,]

where C is an arbitrary constant of integration. Omitting C on an indefinite-integral stem is a classic wrong-option trap; on a definite integral, C cancels and must not appear in the final number.

Linearity: ∫ [αf + βg] dx = α∫f dx + β∫g dx.

Worked power antiderivative. ∫ (6x² − 4x + 5) dx = 6·(x³/3) − 4·(x²/2) + 5x + C = 2x³ − 2x² + 5x + C.

Worked check. Differentiate 2x³ − 2x² + 5x + C to recover 6x² − 4x + 5—always a 5-second self-check on hard items.

Substitution preview (chain in reverse). ∫ 2x(x² + 1)³ dx: let u = x² + 1, du = 2x dx, so ∫ u³ du = u⁴/4 + C = (x² + 1)⁴/4 + C.

Trig / exponential pack (high-frequency). ∫ cos(2x) dx: let u = 2x, du = 2 dx ⇒ (1/2)∫ cos u du = (1/2) sin(2x) + C. Likewise ∫ e^{3x} dx = (1/3)e^{3x} + C. If the stem shows a constant factor inside a composite, pull out the reciprocal of that constant—forgetting it is the most common wrong option.

Initial-condition antiderivative. Suppose dy/dx = 4x − 1 and y(0) = 3. Then y = 2x² − x + C. Plug x = 0: 3 = C, so y = 2x² − x + 3. CAE banks sometimes hide this as “particular solution” wording.

Definite Integrals and the Fundamental Theorem

If F′ = f and f is continuous on [a, b], then

[\int_a^b f(x), dx = F(b) - F(a).]

Interpretation: signed area between y = f(x) and the x-axis from a to b. Regions above the axis contribute positively; below, negatively. For geometric area when f dips below, integrate |f| or split at zeros.

Worked definite integral. Evaluate ∫₀² (3x² − 2x) dx.

Antiderivative F(x) = x³ − x². Then F(2) − F(0) = (8 − 4) − 0 = 4.

Worked with a zero crossing. ∫₋₁² x dx = [x²/2]₋₁² = (4/2) − (1/2) = 2 − 0.5 = 1.5. Signed area: from −1 to 0 the triangle has area 1/2 below (contribute −1/2); from 0 to 2 the triangle has area 2 above; net 2 − 0.5 = 1.5. Pure geometric area of both triangles would be 2.5—read the stem carefully (“integral” vs “area”).

Worked split at a root. Area of the region bounded by y = x² − 1 and the x-axis from x = −1 to x = 2. Roots at x = ±1. On [−1, 1] the curve is below or on the axis; on [1, 2] it is above. Geometric area =

∫₋₁¹ |x² − 1| dx + ∫₁² (x² − 1) dx = ∫₋₁¹ (1 − x²) dx + ∫₁² (x² − 1) dx.

First piece: [x − x³/3]₋₁¹ = (1 − 1/3) − (−1 + 1/3) = 2/3 − (−2/3) = 4/3. Second: [x³/3 − x]₁² = (8/3 − 2) − (1/3 − 1) = (2/3) − (−2/3) = 4/3. Total area = 8/3. The raw signed integral ∫₋₁² (x² − 1) dx equals [x³/3 − x]₋₁² = (8/3 − 2) − (−1/3 + 1) = 2/3 − 2/3 = 0—a trap if the stem asked for area, not net integral.

Properties used on MCQs:

• ∫ₐᵃ f = 0
• ∫ₐᵇ f = −∫ᵇₐ f
• ∫ₐᵇ f + ∫ᵇᶜ f = ∫ₐᶜ f (additivity of intervals)
• If f ≤ g on [a, b], then ∫ₐᵇ f ≤ ∫ₐᵇ g

Area Under a Curve

If f(x) ≥ 0 on [a, b], the area under the curve is exactly ∫ₐᵇ f(x) dx.

Worked area. Area under y = 4 − x² from x = 0 to x = 2: ∫₀² (4 − x²) dx = [4x − x³/3]₀² = 8 − 8/3 = 16/3.

Area between two curves. If y = f(x) is above y = g(x) on [a, b],

[A = \int_a^b \bigl(f(x) - g(x)\bigr), dx.]

Worked between curves. Find the area between y = x and y = x² from their intersection points. Solve x = x² ⇒ x(x − 1) = 0 ⇒ x = 0, 1. On [0, 1], the line is above the parabola:

A = ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6.

Average Value (Quick Extra)

The average value of continuous f on [a, b] is

[f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x), dx.]

For f(x) = 3x² on [0, 2]: ∫₀² 3x² dx = [x³]₀² = 8, so f_avg = 8/2 = 4.

Displacement vs distance (motion link). If v(t) = 3t² − 2 on [0, 2], displacement = ∫₀² v(t) dt = [t³ − 2t]₀² = 8 − 4 = 4. Distance travelled requires integrating |v| and splitting where v = 0 (solve 3t² − 2 = 0 ⇒ t = √(2/3) in the interval)—only attempt that if the stem explicitly says “distance” or “total path.”

On the Exam

Typical stems: “∫ (2x − 1) dx = ?”, “∫₁³ (2x) dx = ?”, “area under y = … from a to b”, and “area between y = x and y = x²”. Prefer evaluating F(b) − F(a) over drawing unless a sign/area distinction is the point of the question. Remember: indefinite → +C; definite → a number. When options include both a signed net and a larger positive “area,” match the wording exactly before selecting.