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)
Adding (i) and (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).
