Definite integrals and area for NDA

Definite integrals are where NDA rewards the student who knows the properties. A question that takes three minutes by brute force often takes twenty seconds once you spot that the function is odd, or that the a + b − x substitution folds the integral onto itself. The syllabus asks for evaluation of definite integrals and the areas of plane regions bounded by curves, and both come up in most papers.

In this post, ∫[a, b] f(x) dx means the integral of f(x) from x = a to x = b.

Typical question types are:

• straight evaluation using an antiderivative;
• a symmetric-limit integral that collapses to 0 by the odd-function rule;
• an integral that looks impossible but yields to the a + b − x property;
• integrals of |x|-type modulus functions or the greatest integer function;
• differentiating an integral whose upper limit is a function of x;
• area under a curve, between a curve and a line, or between two curves.

Evaluating: the fundamental theorem

If F′(x) = f(x), then ∫[a, b] f(x) dx = F(b) − F(a). The constant C cancels, so you never write it.

• ∫[0, 2] x² dx = [x³/3] from 0 to 2 = 8/3.
• ∫[0, π/2] sin x dx = [−cos x] from 0 to π/2 = 0 − (−1) = 1.
• ∫[1, e] (1/x) dx = ln e − ln 1 = 1.

When you substitute u = g(x), change the limits too. For ∫[0, 1] 2x exp(x²) dx (exp(u) means eᵘ), put u = x²: the limits become 0 and 1, and the integral is ∫[0, 1] eᵘ du = e − 1.

The properties that save time

Odd functions include x, x³, sin x, tan x and x cos x. Even functions include x², cos x, |x| and x sin x. A sum like x⁵ + x³ + 1 is odd plus even, so over [−1, 1] only the constant survives.

The a + b − x property in action

Let I = ∫[0, π/2] sin x/(sin x + cos x) dx. Replacing x by π/2 − x swaps sin and cos, so I = ∫[0, π/2] cos x/(cos x + sin x) dx as well. Add the two versions: 2I = ∫[0, π/2] 1 dx = π/2, so I = π/4. The same argument gives π/4 for any power, such as √(sin x)/(√(sin x) + √(cos x)).

Modulus and greatest integer functions

Split the interval where the expression inside changes sign. For ∫[0, 3] |x − 1| dx, the modulus flips at x = 1: ∫[0, 1] (1 − x) dx + ∫[1, 3] (x − 1) dx = 1/2 + 2 = 5/2.

For the greatest integer function ⌊x⌋ (the largest integer not exceeding x), split at the integers. On [0, 1) it is 0 and on [1, 2) it is 1, so ∫[0, 2] ⌊x⌋ dx = 0 + 1 = 1.

Area

Area is always positive; an integral need not be. That single fact decides many questions.

• Area under y = f(x) ≥ 0 from x = a to x = b is ∫[a, b] f(x) dx.
• If the curve goes below the x-axis, take the modulus of each piece separately. ∫[0, 2π] sin x dx = 0, but the area between sin x and the axis over that interval is 2 + 2 = 4.
• Area between two curves is ∫(upper − lower) dx between their points of intersection. Solve the two equations first to find the limits.
• Area with respect to y: for a curve given as x = g(y), use ∫(right − left) dy.

Standard areas worth remembering

The latus-rectum result is a good one to derive once. The upper half of y² = 4ax is y = 2√(ax), and the region is symmetric, so the area is 2∫[0, a] 2√(ax) dx = 4√a × (2/3)a√a = 8a²/3.

Worked NDA-style MCQs

Q1. ∫[0, π/2] √(sin x)/(√(sin x) + √(cos x)) dx equals:
(a) 0 (b) π/2 (c) π/4 (d) 1

Use x → π/2 − x, add the two forms, and 2I = π/2. Answer: (c).

Q2. ∫[−1, 1] (x⁵ + x³ + 1) dx equals:
(a) 0 (b) 1 (c) 2 (d) 4/3

x⁵ and x³ are odd, so their integrals over [−1, 1] vanish. What is left is ∫[−1, 1] 1 dx = 2. Answer: (c).

Q3. ∫[0, 3] |x − 1| dx equals:
(a) 3/2 (b) 5/2 (c) 2 (d) 3

Split at x = 1: 1/2 + 2 = 5/2, as worked above. Answer: (b).

Q4. The area bounded by y = x² and the line y = 4 is:
(a) 16/3 (b) 32/3 (c) 8 (d) 64/3

They meet at x = ±2. The line is above the parabola, so the area is ∫[−2, 2] (4 − x²) dx = 16 − 16/3 = 32/3. Answer: (b).

Q5. The area of the ellipse x²/9 + y²/4 = 1 is:
(a) 6π (b) 13π (c) 36π (d) 12π

a = 3 and b = 2, so πab = 6π. Answer: (a).

Q6. ∫[0, π] x sin x dx equals:
(a) 0 (b) 1 (c) π/2 (d) π

By parts: [−x cos x] from 0 to π, plus ∫[0, π] cos x dx. The first part is −π cos π − 0 = π and the second is sin π − sin 0 = 0. Answer: (d).

Common mistakes

• Treating an integral as an area when the curve crosses the axis. Split at the crossing and add the magnitudes.
• Forgetting to change the limits after a substitution.
• Calling a function odd without checking. x sin x is even, not odd: both factors change sign.
• Subtracting the wrong way in area between curves. Sketch for five seconds to see which curve is on top.
• Missing the chain factor in the Leibniz rule: d/dx ∫[0, x²] cos t dt = 2x cos(x²), not cos(x²).

Practice set

1. ∫[0, 1] x eˣ dx = (a) e (b) 1 (c) e − 1 (d) 2
2. ∫[0, π/4] tan²x dx = (a) 1 (b) π/4 (c) 1 − π/4 (d) π/4 − 1
3. ∫[0, 2] ⌊x⌋ dx, where ⌊x⌋ is the greatest integer function, = (a) 0 (b) 1 (c) 2 (d) 3
4. The area between y = x³, the x-axis and the lines x = −1 and x = 1 is: (a) 0 (b) 1/4 (c) 1/2 (d) 1
5. ∫[0, π/2] cos x/(sin x + cos x) dx = (a) π/2 (b) π/4 (c) 1 (d) 0
6. The area bounded by y² = 4x and the line x = 1 is: (a) 4/3 (b) 8/3 (c) 16/3 (d) 2
7. d/dx ∫[0, x²] cos t dt = (a) cos x² (b) 2x cos(x²) (c) sin(x²) (d) 2x sin(x²)
8. The area enclosed between y = x and y = x³ in the first quadrant is: (a) 1/2 (b) 1/3 (c) 1/4 (d) 1/6

Answers:

1. (b). The antiderivative is eˣ(x − 1); at 1 it is 0, at 0 it is −1, so 0 − (−1) = 1.
2. (c). tan²x = sec²x − 1, so [tan x − x] from 0 to π/4 = 1 − π/4.
3. (b). 0 on [0, 1) and 1 on [1, 2).
4. (c). The integral is 0, but the area is 2∫[0, 1] x³ dx = 2 × 1/4 = 1/2.
5. (b). The a + b − x property, as in Q1.
6. (b). Here a = 1, so 8a²/3 = 8/3.
7. (b). Leibniz rule: cos(x²) × 2x.
8. (c). They meet at x = 0 and 1, with x above x³: 1/2 − 1/4 = 1/4.

What to do next

• Copy the properties table and, beside each row, write one example you solved yourself.
• Derive the 8a²/3 and 16a²/3 results once; the second needs the intersection points (0, 0) and (4a, 4a).
• If an antiderivative held you up, go back to indefinite integration. Next in the calculus run is differential equations.