Definite integral and the fundamental theorem
The distance covered between 2 pm and 5 pm is the total of speed × tiny time slices in that window only; a definite integral gives a number for a chosen interval.
If F′ = f then ∫ₐᵇ f(x) dx = F(b) − F(a). It is the signed area between the curve and the x-axis from x = a to x = b. No constant C is needed.
Rule 1: Evaluate — Find an antiderivative, put the upper limit, subtract the value at the lower limit.
Rule 2: Reversal — ∫ₐᵇ f dx = −∫ᵇₐ f dx; ∫ₐᵃ f dx = 0; ∫ₐᵇ f = ∫ₐᶜ f + ∫ᶜᵇ f.
How is the formula obtained? : ∫₀² x² dx = [x³/3]₀² = 8/3 − 0 = 8/3 ≈ 2.667.
- Easy : ∫₀² x² dx = 8/3.
- Medium : ∫₀^{π/2} cos x dx = sin(π/2) − sin 0 = 1.
- Exam-type : ∫₁^e (1/x) dx = log e − log 1 = 1.
The result of a definite integral is a number.
Subtracting in the wrong order (lower − upper).
Substitute, then subtract: upper minus lower. Memory clue : "Top minus bottom."
If f changes sign, the integral is the difference of areas, not the total area.
CET-style example : ∫₀² x² dx = (A) 8/3 (B) 4 (C) 2 (D) 8
Answer : (A). Time-saving method : Use x³/3.
Spot the difference
| Integral | Result |
|---|---|
| Indefinite | a function + C |
| Definite | a number |
Practice questions
- 1∫₀¹ x dx
- 2∫₀² x² dx
- 3∫₀^{π/2} cos x dx
- 4∫₁^e dx/x
- 5∫₀¹ eˣ dx
- 6∫₁² (x + 1/x²) dx
- 7∫₀^π sin x dx
- 8∫₀^1 x eˣ dx
Show answers (try them yourself first)
- 11/2.
- 28/3.
- 31.
- 41.
- 5e − 1 ≈ 1.718.
- 6[x²/2 − 1/x]₁² = (2 − ½) − (½ − 1) = 2.
- 72.
- 8[(x − 1)eˣ]₀¹ = 0 − (−1) = 1.
Substitution in definite integrals
Changing currency also changes the amounts at both ends; in the same way, changing the variable also changes the limits.
In ∫ₐᵇ f(g(x)) g′(x) dx put u = g(x); the limits become g(a) and g(b). Then no back-substitution is needed.
Rule 1: Change limits — x = a ⇒ u = g(a); x = b ⇒ u = g(b).
How is the formula obtained? : ∫₀¹ (2x + 1)³ dx: u = 2x + 1, du = 2dx, limits 1 to 3. ½∫₁³ u³ du = ½ [u⁴/4]₁³ = (81 − 1)/8 = 10. Check: expanding gives 10.
- Easy : ∫₀¹ 2x/(1 + x²) dx: u = 1 + x², limits 1 to 2 ⇒ [log u]₁² = log 2.
- Medium : ∫₀^{√3} x√(x² + 1) dx: u = x² + 1, limits 1 to 4 ⇒ ½·(2/3)[u^{3/2}]₁⁴ = (1/3)(8 − 1) = 7/3.
- Exam-type : ∫₀^{π/2} sin x cos x dx = ½ (u = sin x, limits 0 to 1).
Either change the limits or go back to x; never mix.
Putting old limits on the new variable.
Write the new limits in brackets beside u immediately. Memory clue : "New variable, new limits."
The substitution must be a valid single-valued function on the whole interval.
CET-style example : ∫₀¹ 2x/(1 + x²) dx = (A) log 2 (B) 1 (C) π/4 (D) 2
Answer : (A). Time-saving method : u = 1 + x², limits 1, 2.
Spot the difference
| Method | Limits |
|---|---|
| Change variable | convert limits to u |
| Back-substitute | keep x limits |
Practice questions
- 1∫₀¹ 2x/(1 + x²) dx
- 2∫₀¹ (2x + 1)³ dx
- 3∫₀^{π/2} sin x cos x dx
- 4∫₀¹ x e^{x²} dx
- 5∫₁² 1/(2x + 1) dx
- 6∫₀^{√3} x√(x² + 1) dx
- 7∫₀^{π/4} tan x dx
- 8∫₀¹ x/(x² + 1)² dx
Show answers (try them yourself first)
- 1log 2.
- 210.
- 31/2.
- 4(e − 1)/2.
- 5½ log(5/3) ≈ 0.255.
- 67/3.
- 7[−log cos x]₀^{π/4} = ½ log 2.
- 8[−1/(2(x² + 1))]₀¹ = −¼ + ½ = 1/4.
Properties of definite integrals
A symmetric seesaw shows how matching left and right halves cancel; some integrals can be found without calculating, just by symmetry.
King's property: ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a − x) dx. If f is even, ∫₋ₐᵃ f = 2∫₀ᵃ f; if f is odd, ∫₋ₐᵃ f = 0.
Rule 1: Even / odd — Even: f(−x) = f(x); odd: f(−x) = −f(x).
Rule 2: King's trick — Add I = ∫f(x) and I = ∫f(a − x) to get 2I.
How is the formula obtained? : I = ∫₀^{π/2} sin x/(sin x + cos x) dx. By the king's property I = ∫₀^{π/2} cos x/(cos x + sin x) dx. Adding: 2I = ∫₀^{π/2} 1 dx = π/2, so I = π/4.
- Easy : ∫₋₁¹ x³ dx = 0 (odd function).
- Medium : ∫₋₂² x² dx = 2∫₀² x² dx = 16/3.
- Exam-type : ∫₀^{π/2} sin x/(sin x + cos x) dx = π/4.
Check even or odd before integrating.
Claiming ∫₋ₐᵃ f = 0 for an even function.
Odd function over symmetric limits: answer 0. Memory clue : "Odd cancels, even doubles."
x⁴ is even, x³ is odd; the sign of the exponent matters for cos x vs sin x too.
CET-style example : ∫₋₁¹ x³ dx = (A) 0 (B) 1/2 (C) 2 (D) 1/4
Answer : (A). Time-saving method : x³ is odd.
Spot the difference
| f | Over [−a, a] |
|---|---|
| Even | 2∫₀ᵃ f |
| Odd | 0 |
Practice questions
- 1∫₋₁¹ x³ dx
- 2∫₋₂² x² dx
- 3∫₋π^π sin x dx
- 4∫₋π/2^{π/2} cos x dx
- 5∫₀^{π/2} sin x/(sin x + cos x) dx
- 6∫₀^{π/2} cos x/(sin x + cos x) dx
- 7∫₀¹ x(1 − x) dx using king's property
- 8∫₀^{π} sin²x dx
Show answers (try them yourself first)
- 10.
- 216/3.
- 30.
- 42.
- 5π/4.
- 6π/4.
- 7∫₀¹ (1 − x)x dx; direct: ½ − ⅓ = 1/6.
- 8π/2.