# Properties and Formulas for Definite Integrals

Here, you will learn formulas for definite integrals and properties of definite integrals with examples.

Let’s begin –

A definite integral is denoted by $$\int_{a}^{b}$$ f(x)dx which represent the algebraic area bounded by the curve y = f(x), the ordinates x = a, x = b and the x-axis.

## Properties and Formulas for Definite Integrals

(a)  $$\int_{a}^{b}$$ f(x)dx = $$\int_{a}^{b}$$ f(t)dt provided f is same

(b)  $$\int_{a}^{b}$$ f(x)dx = – $$\int_{b}^{a}$$ f(x)dx

(c)  $$\int_{a}^{b}$$ f(x)dx = $$\int_{a}^{c}$$ f(x)dx + $$\int_{c}^{b}$$ f(x)dx , where c may lie inside or outside the interval [a,b]. This property is to be used when f is piecewise continous in (a, b).

(d)  $$\int_{a}^{a}$$ f(x)dx = $$\int_{0}^{a}$$ [f(x) + f(-x)]dx = $$\begin{cases} 0 & \text{if f(x) is an odd function}\ \\ 2\int_{a}^{b} f(x)dx & \text{if f(x) is an even function}\ \end{cases}$$

Example : Evaluate $$\int_{1/2}^{1/2}$$ $$cosx ln{({1+x\over 1-x})}$$ dx

Solution : f(-x) = $$cos(-x) ln{({1-x\over 1+x})}$$ = – $$cosx ln{({1+x\over 1-x})}$$ = f(-x)

$$\implies$$   f(x) is odd

Hence, the value of the given interval is 0.

(e)  $$\int_{a}^{b}$$ f(x)dx = $$\int_{a}^{b}$$ f(a+b-x)dx, In particular $$\int_{0}^{a}$$ f(x)dx = $$\int_{0}^{a}$$ f(a-x)dx

Example : Evaluate $$\int_{0}^{\pi/2}$$ $$asinx+bcosx\over sinx+cosx$$ dx

Solution : I = $$\int_{0}^{\pi/2}$$ $$asinx+bcosx\over sinx+cosx$$ dx     ….(i)

I = $$\int_{0}^{\pi/2}$$ $$asin(\pi/2-x)+bcos(\pi/2-x)\over sin(\pi/2-x)+cos(\pi/2-x)$$ dx = $$\int_{0}^{\pi/2}$$ $$acosx+bsinx\over sinx+cosx$$ dx     ….(ii)

2I = $$\int_{0}^{\pi/2}$$ $$a+b)(sinx+cosx)\over sinx+cosx$$ dx = $$\int_{0}^{\pi/2}$$ (a+b) dx = (a+b)$$\pi/2$$
$$\implies$$   I = (a+b)$$\pi/4$$
(f)  $$\int_{0}^{2a}$$ f(x)dx = $$\int_{0}^{a}$$ f(x)dx + $$\int_{0}^{a}$$ f(2a-x)dx = $$\begin{cases} 2\int_{0}^{a} f(x)dx & \text{if}\ f(2a-x) = f(x) \\ 0 & \text{if}\ f(2a-x) = -f(x) \end{cases}$$.
(g)  $$\int_{0}^{nT}$$ f(x)dx = n$$\int_{0}^{T}$$ f(x)dx, (n $$\in$$ I); where T is the period of the function i.e. f(T+x) = f(x)
(h)  $$\int_{a+nT}^{b+nT}$$ f(x)dx = $$\int_{a}^{b}$$ f(x)dx, where f(x) is periodic with period T & n $$\in$$ I.
(i)  $$\int_{mT}^{nT}$$ f(x)dx = (n-m)$$\int_{0}^{T}$$ f(x)dx, where f(x) is periodic with period T & (n, m $$\in$$ I).