Integral Calculus

Indefinite integrals, definite integrals, properties, areas under curves, applications.

Indefinite integrals

Standard forms, substitution, by parts.

Integration techniques — substitution, by parts, partial fractions
Notes

Three methods cover 90% of JEE integral problems. Pattern recognition is everything.

1. Substitution (u-sub). When the integrand contains f(g(x)) · g'(x), let u = g(x).

Example: ∫ 2x · cos(x²) dx. Let u = x², du = 2x dx. Becomes ∫ cos u du = sin u + C = sin(x²) + C.

Common patterns:

  • ∫ f(ax+b) dx → u = ax+b
  • ∫ x · f(x²) dx → u = x²
  • ∫ f(sin x) cos x dx → u = sin x
  • ∫ tan x dx = ∫ sin x / cos x dx → u = cos x → −ln|cos x| + C

2. Integration by parts (LIATE rule).

∫ u dv = uv − ∫ v du

Pick u in this order (LIATE):

  • Logarithmic
  • Inverse trig
  • Algebraic
  • Trigonometric
  • Exponential

Example: ∫ x · ln x dx. ln x is L (higher in LIATE) → u = ln x, dv = x dx.
du = dx/x, v = x²/2. Result: (x²/2) ln x − ∫ (x²/2)(1/x) dx = (x²/2) ln x − x²/4 + C.

Special trick: ∫ eˣ [f(x) + f'(x)] dx = eˣ · f(x) + C.

3. Partial fractions for rational integrands P(x)/Q(x) where deg(P) < deg(Q).

Example: ∫ 1/[(x−1)(x+2)] dx.
Let 1/[(x−1)(x+2)] = A/(x−1) + B/(x+2). Multiply: 1 = A(x+2) + B(x−1).
x=1: 1 = 3A → A = 1/3. x=−2: 1 = −3B → B = −1/3.
Integral = (1/3) ln|x−1| − (1/3) ln|x+2| + C = (1/3) ln|(x−1)/(x+2)| + C.

Standard integrals to memorize:

  • ∫ dx/(x² + a²) = (1/a) tan⁻¹(x/a) + C
  • ∫ dx/(x² − a²) = (1/2a) ln|(x−a)/(x+a)| + C
  • ∫ dx/√(a² − x²) = sin⁻¹(x/a) + C
  • ∫ dx/√(x² + a²) = ln|x + √(x² + a²)| + C
  • ∫ dx/√(x² − a²) = ln|x + √(x² − a²)| + C

Properties of definite integrals (often more useful than computing):

  • ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a−x) dx (king property)
  • ∫₋ₐᵃ f(x) dx = 2∫₀ᵃ f(x) dx if f is even; 0 if odd.
  • ∫₀^(2a) f(x) dx = 2∫₀ᵃ f(x) dx if f(2a−x) = f(x); 0 if = −f(x).

Definite integrals and properties

Fundamental theorem, properties.

Definite integrals — properties and applications to areas
Notes

Fundamental theorem of calculus:

∫_a^b f(x) dx = F(b) − F(a)

where F is any antiderivative of f.


PROPERTIES OF DEFINITE INTEGRALS (the "King property" and friends):

1. ∫_a^b f(x) dx = −∫_b^a f(x) dx (reverse limits → negate).

2. ∫_a^a f(x) dx = 0.

3. ∫_a^b f(x) dx = ∫_a^c f(x) dx + ∫_c^b f(x) dx (split at c).

4. King property: ∫_a^b f(x) dx = ∫_a^b f(a + b − x) dx.

This is the magic property — often simplifies integrals dramatically.

5. ∫_0^a f(x) dx = ∫_0^a f(a − x) dx (special case of #4 with a = 0).

6. For even/odd functions:

  • f(x) even: ∫_{−a}^a f(x) dx = 2∫_0^a f(x) dx.
  • f(x) odd: ∫_{−a}^a f(x) dx = 0.

7. Period property: if f has period T,
∫_0^{nT} f(x) dx = n × ∫_0^T f(x) dx.

8. ∫_0^{2a} f(x) dx = 2∫_0^a f(x) dx if f(2a − x) = f(x), and = 0 if f(2a − x) = −f(x).


Worked example using King property:

Find I = ∫_0^{π/2} sin x / (sin x + cos x) dx.

Using property #4 with a + b − x = π/2 − x:
I = ∫_0^{π/2} sin(π/2 − x) / (sin(π/2 − x) + cos(π/2 − x)) dx = ∫_0^{π/2} cos x / (cos x + sin x) dx.

Add the two: 2I = ∫_0^{π/2} (sin x + cos x)/(sin x + cos x) dx = ∫_0^{π/2} 1 dx = π/2.

I = π/4.


Worked example using odd function property:

∫_{−π}^{π} sin³x dx.

sin³x is odd (since sin is odd, odd³ = odd). So integral over symmetric interval = 0.


AREA UNDER A CURVE

Area between f(x) and x-axis from x = a to x = b:

A = ∫_a^b |f(x)| dx

(Take absolute value because area is always positive; integral can be negative if f < 0.)

If f changes sign in [a, b], split the integral at zeros of f.


Area between two curves y = f(x) (top) and y = g(x) (bottom):

A = ∫_a^b [f(x) − g(x)] dx

where a, b are the x-coordinates of intersections.


Area bounded by curve x = f(y), y-axis, y = c, y = d:

A = ∫_c^d f(y) dy

(integrate horizontally if it's simpler).


Worked example: Area of a circle of radius r.

The upper half is y = √(r² − x²) for x ∈ [−r, r].

Total area = 2 × ∫_{−r}^r √(r² − x²) dx.

Using x = r sin θ, dx = r cos θ dθ:
= 2 × ∫{−π/2}^{π/2} r cos θ × r cos θ dθ = 2r² × ∫{−π/2}^{π/2} cos²θ dθ
= 2r² × ∫_{−π/2}^{π/2} (1 + cos 2θ)/2 dθ
= 2r² × π/2 = πr². ✓


Area of an ellipse x²/a² + y²/b² = 1: πab (similar derivation).

Area enclosed by parabola y² = 4ax and its latus rectum: (8/3)a².

Common JEE/NEET-style problems:

  1. Area between y = x² and y = x.
  2. Area between y = sin x and y = cos x in [0, π].
  3. Area between two parabolas / parabola and line / ellipse and line.

AVERAGE VALUE OF A FUNCTION over [a, b]:

f_avg = (1/(b − a)) × ∫_a^b f(x) dx

Useful in physics (average velocity, average current).


INTEGRATION BY PARTS (useful for definite integrals too):

∫_a^b u dv = [uv]_a^b − ∫_a^b v du

Apply LIATE rule (covered in Pack 5).


WALLIS' FORMULA:

∫_0^{π/2} sin^n x dx = ∫_0^{π/2} cos^n x dx

For even n: = [(n−1)/n] × [(n−3)/(n−2)] × ... × (1/2) × (π/2).
For odd n: = [(n−1)/n] × [(n−3)/(n−2)] × ... × (2/3).

Saves you from doing integration by parts repeatedly.

Areas under curves

Area between curves, applications.

No published notes for this topic yet.