if you draw a curve on a graph and want the area trapped between the curve and the x-axis from point a to point b, the definite integral does exactly that job
Integral from a to b of f(x) dx = F(b) - F(a), where F(x) is the antiderivative of f(x). Here, 'a' is the lower limit and 'b' is the upper limit. --- KEY RULES / PROPERTIES --- 1
Integral from a to b = -(Integral from b to a) 2. Same limits = zero: Integral from a to a of f(x) dx = 0 3
Integral from a to b = Integral from a to c + Integral from c to b (where a < c < b) 4
If f(-x) = f(x), then Integral from -a to a = 2 × Integral from 0 to a 5
If f(-x) = -f(x), then Integral from -a to a = 0 6. King's Property (MOST IMPORTANT for NDA): Integral from a to b of f(x) dx = Integral from a to b of f(a+b-x) dx ---
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--- NDA regularly asks: - Direct evaluation of simple definite integrals (substitution based) - Questions using King's Property to simplify tricky integrals - Even/Odd function questions with symmetric limits - Area under a parabola, line, or circle --- SHORTCUT / TRICK --- TRICK 1 (King's Property): When you see Integral from 0 to pi of [x * sin(x)] or similar, apply King's Property. Replace x with (a+b-x). Add the original and new form. The tricky part cancels out and you solve easily.
This saves 3-4 minutes in the exam. TRICK 2 (Even-Odd Quick Check): If the limits are -a to +a, always check if the function is odd. If yes, answer is ZERO. No calculation needed. NDA uses this trap often. ---
Let I = Integral[0 to pi] x/(1 + sin x) dx
Apply King's Property. Replace x with (0 + pi - x) = (pi - x) New form: I = Integral[0 to pi] (pi - x)/(1 + sin(pi - x)) dx Note: sin(pi - x) = sin x, so denominator stays same. New form: I = Integral[0 to pi] (pi - x)/(1 + sin x) dx
Add both forms of I: 2I = Integral[0 to pi] [x + pi - x]/(1 + sin x) dx 2I = Integral[0 to pi] pi/(1 + sin x) dx 2I = pi × Integral[0 to pi] 1/(1 + sin x) dx
Multiply top and bottom by (1 - sin x): 1/(1 + sin x) = (1 - sin x)/(1 - sin²x) = (1 - sin x)/cos²x = sec²x - sec x tan x
Integrate: Integral of sec²x = tan x, Integral of sec x tan x = sec x Evaluate from 0 to pi carefully using limits → Result = 2 So 2I = pi × 2 = 2pi Final Answer: I = pi ---
--- Students forget that area can NEVER be negative. If the curve goes below the x-axis, the integral gives a negative value. Always take the absolute value (modulus) when finding area.
Also, never confuse the definite integral VALUE with AREA — they are equal only when f(x) is above the x-axis.
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