Definition of Definite Integral
Unlike the indefinite integral, a definite integral has specific lower and upper limits \(a\) and \(b\), and evaluates to a real number (not a family of functions).
where \(F(x)\) is any antiderivative of \(f(x)\), i.e., \(F'(x)=f(x)\).
The definite integral is formally defined as the limit of a Riemann sum — the area under the curve approximated by \(n\) rectangles as \(n \to \infty\).
Here \(h\) is the width of each rectangle, and \(f(a+rh)\) is the height of the \(r\)-th rectangle.
Useful Summation Formulas (needed for limit-of-sum method):
\(\displaystyle\sum_{r=1}^{n}r = \frac{n(n+1)}{2}\)
\(\displaystyle\sum_{r=1}^{n}r^2 = \frac{n(n+1)(2n+1)}{6}\)
\(\displaystyle\sum_{r=1}^{n}r^3 = \left[\frac{n(n+1)}{2}\right]^2\)
Fundamental Theorem of Calculus
If \(f\) is continuous on \([a,b]\), define \(g(x) = \displaystyle\int_a^x f(t)\,dt\). Then \(g\) is differentiable and:
This says: differentiation and integration are inverse operations.
If \(F\) is any antiderivative of \(f\) on \([a,b]\), then:
This is the key result used for evaluating all definite integrals in the board exam.
Evaluate \(\int f(x)\,dx\) (without \(+C\) — the constant cancels out).
Compute \(F(b) - F(a)\) where \(b\) is the upper limit and \(a\) is the lower limit.
Simplify the result. The answer is a real number, not a function.
Properties of Definite Integrals
These 8 properties are the backbone of definite integration — they let you simplify, flip, and evaluate difficult integrals without computing them directly.
Swapping the upper and lower limits changes the sign of the integral.
When both limits are equal, the integral is zero.
The integral can be split at any interior point \(c\) between \(a\) and \(b\).
Useful when \(f(x)\) has different expressions on sub-intervals (e.g., modulus functions).
The value of a definite integral is independent of the variable of integration.
Replace \(x\) by \((a+b-x)\) — the value of the integral does not change.
★ Most frequently used property in CBSE/JEE exams. Used to prove many results.
For integrals from \(0\) to \(2a\):
For a symmetric interval \([-a, a]\):
If \(f\) is periodic with period \(T\) [i.e., \(f(x+T)=f(x)\)], then:
Worked Examples Using Properties
Evaluate \(\displaystyle I = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}\,dx\)
Apply P₅ with \(a=0,\; b=\pi/2\): replace \(x \to \frac{\pi}{2}-x\)
Adding the two expressions for \(I\):
Evaluate \(\displaystyle\int_{-2}^{2} x^3\,dx\)
Since \(f(x)=x^3\) is an odd function (because \(f(-x) = -x^3 = -f(x)\)) and the interval is symmetric:
Evaluate \(\displaystyle\int_{-1}^{2} |x|\,dx\)
Since \(|x| = -x\) for \(x < 0\) and \(|x|=x\) for \(x \geq 0\), split at \(0\):
Evaluate \(\displaystyle\int_0^1 \frac{2x}{1+x^2}\,dx\)
Let \(t = 1+x^2 \Rightarrow dt = 2x\,dx\). Change limits: \(x=0 \Rightarrow t=1;\; x=1 \Rightarrow t=2\)
Key Rule: When you substitute in a definite integral, always change the limits of integration to match the new variable — do not back-substitute.
Properties — Quick Reference
| Prop | Name | Statement |
|---|---|---|
| P₁ | Reverse Limits | \(\displaystyle\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx\) |
| P₂ | Zero Width | \(\displaystyle\int_a^a f(x)\,dx = 0\) |
| P₃ | Additivity | \(\displaystyle\int_a^b = \int_a^c + \int_c^b\) |
| P₄ | Dummy Variable | \(\displaystyle\int_a^b f(x)\,dx = \int_a^b f(t)\,dt\) |
| P₅ ⭐ | King's Property | \(\displaystyle\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx\) |
| P₆ | Half-Interval | \(\displaystyle\int_0^{2a} f(x)\,dx = \int_0^a [f(x)+f(2a-x)]\,dx\) |
| P₇ | Even / Odd | \(\displaystyle\int_{-a}^{a} f\,dx = \begin{cases}2\int_0^a f\,dx & \text{even}\\0 & \text{odd}\end{cases}\) |
| P₈ | Periodic | \(\displaystyle\int_0^{nT} f(x)\,dx = n\int_0^T f(x)\,dx\) |
Board Exam Strategy: P₅ (King's Property) is the single most tested property — it appears in at least one 5-mark question every year. Whenever you see \(\sin\)/\(\cos\) or \(\ln\) in the integrand with symmetric limits, try P₅ first. P₇ (Even/Odd) can save you from long calculations on \([-a,a]\) type problems.