In calculus, integration is a very important part of the computation. It is used for many problem-solving approaches in other areas like Physics and Chemistry. Sometimes we need to compute integral with a definite range of values. These are called Definite integrals. The definite integral formula is applicable to the upper and lower limits given. In this article, we will discuss the Definite Integral Formula. Let us learn this interesting concept!
Definite Integral Formula
Concept of Definite Integrals
The definite integral is defined as the limit and summation that we looked at in the last section to find the net area between the given function and the x-axis. Here note that the notation for the definite integral is very similar to the notation for an indefinite integral. One example of this is as given:
\(\large \int_{a}^{\infty}f(x)dx=\lim_{b\rightarrow \infty}\left [ \int_{a}^{b}f(x)dx\right ]\)
\(\large \int_{a}^{b}f(x)dx=F(b)-F(a)\)
Here, F(a) is the lower limit value of the integral and F(b) is the upper limit value of the integral.
There is also a little bit of terminology that we can get out of the way. The number a at the bottom of the integral sign is called the lower limit and the number b at the top of the integral sign is called the upper limit. Although a and b were given as an interval the lower limit does not necessarily need to be smaller than the upper limit. Thus we will often call a and b as the interval of integration.
Some Properties:
- \(\displaystyle \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} =Â –Â \int_{{\,b}}^{{\,a}}{{f\left( x \right)\,dx}}\)
- \(\displaystyle \int_{{\,a}}^{{\,a}}{{f\left( x \right)\,dx}} = 0\)
- \(\displaystyle \int_{{\,a}}^{{\,b}}{{cf\left( x \right)\,dx}} = c\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}}\)
- \(\displaystyle \int_{{\,a}}^{{\,b}}{{f\left( x \right) \pm g\left( x \right)\,dx}} = \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} \pm \int_{{\,a}}^{{\,b}}{{g\left( x \right)\,dx}}\)
- \(\displaystyle \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} = \int_{{\,a}}^{{\,c}}{{f\left( x \right)\,dx}} + \int_{{\,c}}^{{\,b}}{{f\left( x \right)\,dx}}\)
- \(\displaystyle \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} = \int_{{\,a}}^{{\,b}}{{f\left( t \right)\,dt}}\)
Some Popular Formulas
For Rational or Irrational Expression
- \(\large \int_{a}^{\infty }\frac{dx}{x^{2}+a^{2}}=\frac{\pi }{2a}\)
- \(\large \int_{a}^{\infty }\frac{x^{m}dx}{x^{n}+a^{n}}=\frac{\pi a^{m-n+1}}{n\sin \left ( \frac{(m+1)\pi }{n} \right )},0< m+1< n \\\)
- \(\large \int_{a}^{\infty }\frac{x^{p-1}dx}{1+x}=\frac{\pi }{\sin (p\pi )},0< p< 1Â \\\)
- \(\large \int_{a}^{\infty }\frac{dx}{\sqrt{a^{2}-x^{2}}}=\frac{\pi }{2} \\\)
- \(\large \int_{a}^{\infty }\sqrt{a^{2}-x^{2}}dx=\frac{\pi a^{2}}{4}Â \\\)
For Trigonometric Functions
- \(\large \int_{0}^{\pi }\sin(mx)\sin (nx)dx=\left\{\begin{matrix} 0 & if\;m\neq n\\ \frac{\pi }{2} & if\;m=n \end{matrix}\right.\;m,n\;positive\;integers \\\)
- \(\large \int_{0}^{\pi }\cos (mx)\cos (nx)dx=\left\{\begin{matrix} 0 & if\;m\neq n\\ \frac{\pi }{2} & if\;m=n \end{matrix}\right.\;m,n\;positive\;integers \\\)
- \(\large \int_{0}^{\pi }\sin (mx)\cos (nx)dx=\left\{\begin{matrix} 0 & if\;m+n\;even\\ \frac{2m}{m^{2}-n^{2}} & if\;m+n\;odd \end{matrix}\right.\;m,n\;integers \\\)
Solved Examples for Definite Integral Formula
Q.1: Find the value of definite integral: \(\displaystyle \int_{{\,2}}^{{\,0}}{{{x^2} + 1\,dx}}\)
Solution: In this case we can use the property to get:
\(\begin{align*}\int_{{\,2}}^{{\,0}}{{{x^2} + 1\,dx}} & =Â – \int_{{\,0}}^{{\,2}}{{{x^2} + 1\,dx}}\\ &Â =Â – \frac{{14}}{3}\end{align*}\)
Q2:Â Given that: \(\displaystyle \int_{{\,6}}^{{\, – 10}}{{f\left( x \right)\,dx}} = 23 \) &
\(\displaystyle \int_{{\, – 10}}^{{\,6}}{{g\left( x \right)\,dx}} =Â – 9\)
Determine the value of: \(\int_{{\, – 10}}^{{\,6}}{{2f\left( x \right)\, – 10g\left( x \right)dx}}\)
Solution: We will first break up the integral using property and then to factor out the constants.
\(\begin{align*}\int_{{\, – 10}}^{{\,6}}{{2f\left( x \right)\, – 10g\left( x \right)dx}} & = \int_{{\, – 10}}^{{\,6}}{{2f\left( x \right)\,dx}} – \int_{{\, – 10}}^{{\,6}}{{10g\left( x \right)dx}}\\ &Â = 2\int_{{\, – 10}}^{{\,6}}{{f\left( x \right)\,dx}} – 10\int_{{\, – 10}}^{{\,6}}{{g\left( x \right)dx}}\end{align*}\)
Since the limits on the first integral are interchanged we will add a minus sign of course. After it we can plug in the known values of the integrals.
\(\begin{align*}\int_{{\, – 10}}^{{\,6}}{{2f\left( x \right)\, – 10g\left( x \right)dx}} & =Â – 2\int_{{\,6}}^{{\, – 10}}{{f\left( x \right)\,dx}} – 10\int_{{\, – 10}}^{{\,6}}{{g\left( x \right)dx}}\\ &Â =Â – 2\left( {23} \right) – 10\left( { – 9} \right)\\ &Â = 44\end{align*}\)
I get a different answer for first example.
I got Q1 as 20.5
median 23 and
Q3 26