JEE Main & Advanced Mathematics Definite Integration Properties of Definite Integral

Properties of Definite Integral

Category : JEE Main & Advanced

(1) \[\int_{a}^{b}{f(x)dx}=\int_{a}^{b}{f(t)\,dt}\] i.e., The value of a definite integral remains unchanged if its variable is replaced by any other symbol.

 

 

(2) \[\int_{a}^{b}{f(x)dx=-\int_{b}^{a}{f(x)dx}}\] i.e., by the interchange in the limits of definite integral, the sign of the integral is changed.

 

 

(3) \[\int_{a}^{b}{f(x)dx=\int_{a}^{c}{f(x)dx}+\int_{c}^{b}{f(x)dx}}\],   (where \[a<c<b\]) 

 

 

or \[\int_{a}^{b}{f(x)dx}=\int_{a}^{{{c}_{1}}}{f(x)dx}+\int_{{{c}_{1}}}^{{{c}_{2}}}{f(x)dx+.....+\int_{{{c}_{n}}}^{b}{f(x)dx;}}\] (where \[a<{{c}_{1}}<{{c}_{2}}<........{{c}_{n}}<b\])

 

 

Generally this property is used when the integrand has two or more rules in the integration interval.

 

 

This is useful when \[f\,(x)\] is not continuous in \[[a,\,\,b]\] because we can break up the integral into several integrals at the points of discontinuity so that the function is continuous in the sub-intervals.

 

 

(4) \[\int_{0}^{a}{f(x)dx=\int_{0}^{a}{f(a-x)dx}}\] : This property can be used only when lower limit is zero. It is generally used for those complicated integrals whose denominators are unchanged when \[x\] is replaced by \[(a-x)\].

 

 

Following integrals can be obtained with the help of above property.

 

 

(i) \[\int_{0}^{\pi /2}{\frac{{{\sin }^{n}}x}{{{\sin }^{n}}x+{{\cos }^{n}}x}}\,dx=\int_{0}^{\pi /2}{\frac{{{\cos }^{n}}x}{{{\cos }^{n}}x+{{\sin }^{n}}x}dx=\frac{\pi }{4}}\]    

 

 

(ii) \[\int_{0}^{\pi /2}{\frac{{{\tan }^{n}}x}{1+{{\tan }^{n}}x}dx=\int_{0}^{\pi /2}{\frac{{{\cot }^{n}}x}{1+{{\cot }^{n}}x}dx=\frac{\pi }{4}}}\]

 

 

(iii) \[\int_{0}^{\pi /2}{\frac{1}{1+{{\tan }^{n}}x}dx=\int_{0}^{\pi /2}{\frac{1}{1+{{\cot }^{n}}x}}dx=\frac{\pi }{4}}\]      

 

(iv) \[\int_{0}^{\pi /2}{\frac{{{\sec }^{n\,}}x}{{{\sec }^{n}}\,x+\text{cose}{{\text{c}}^{n}}x}\,dx=}\int_{0}^{\pi /2}{\,\,\,\,}\frac{\text{cose}{{\text{c}}^{n\,}}x}{\text{cose}{{\text{c}}^{n}}\,x+{{\sec }^{n}}x}\,dx=\frac{\pi }{4}\]

 

 

(v) \[\int_{0}^{\pi /2}{\,\,\,f(\sin 2x)\sin xdx=}\int_{0}^{\pi /2}{\,\,\,f(\sin 2x)\cos xdx}\]

 

 

(vi) \[\int_{0}^{\pi /2}{f(\sin x)dx=\int_{0}^{\pi /2}{\,\,\,\,f(\cos x)dx}}\]         

 

 

(vii) \[\int_{0}^{\pi /4}{\log (1+\tan x)dx=\frac{\pi }{8}\log 2}\]        

 

 

(viii) \[\int_{0}^{\pi /2}{\,\,\,\,\,\log \sin xdx}=\int_{0}^{\pi /2}{\,\,\,\,\,\log \cos xdx}=\frac{-\pi }{2}\log 2=\frac{\pi }{2}\log \frac{1}{2}\]

 

 

(ix) \[\int_{0}^{\pi /2}{{}}\frac{a\,\sin \,x+b\,\cos \,x}{\sin \,x+\cos \,x}\,dx=\int_{0}^{\pi /2}{{}}\frac{a\,\sec \,x+b\,\text{cosec}\,x}{\sec \,x+\text{cosec}\,x}\,dx\]\[\int_{0}^{\pi /2}{{}}\frac{a\,\sin \,x+b\,\cos \,x}{\sin \,x+\cos \,x}\,dx=\int_{0}^{\pi /2}{{}}\frac{a\,\sec \,x+b\,\cos ec\,x}{\sec \,x+\cos ec\,x}\,dx=\int_{0}^{\pi /2}{{}}\frac{a\,\tan \,x+b\,\cot \,x}{\tan \,x+\cot \,x}\,dx=\frac{\pi }{4}(a+b)\]

 

 

(5) \[\int_{-a}^{a}{{}}f(x)\,dx=\int_{0}^{a}{\text{ }\!\![\!\!\text{ }f(x)\,+f(-x)\text{ }\!\!]\!\!\text{ }}\text{ }dx\].

 

 

In special case :

 

 

\[\int_{-a}^{a}{f(x)\,dx}=\left\{ \begin{array}{*{35}{l}} 2\int_{0}^{a}{f(x)\,dx},\,\,\text{if}\,f(x)\,\,\text{is}\,\,\text{even function or }f(-x)=f(x)\text{ }  \\ \,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\text{if}\,\,f(x)\,\,\text{is odd function or }f(-x)=-f(x)  \\\end{array} \right.\]  

 

 

This property is generally used when integrand is either even or odd function of \[x\].

 

 

(6) \[\int_{0}^{2a}{\,\,f(x)dx}=\int_{0}^{a}{f(x)dx+\int_{0}^{a}{{}}\text{ }f(2a-x)\,dx}\]

 

 

In particular, \[\int_{0}^{2a}{{}}\,f(x)\,dx\,=\,\left\{ \begin{array}{*{35}{l}} 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\text{if}\,\,\,\,f(2a- x)=-f(x)  \\ 2\int_{0}^{a}{f(x)\,dx}\,,\,\,\,\text{if}\,\,\,\,\,f(2a-x)=f(x)  \\ \end{array} \right.\]

 

 

It is generally used to make half the upper limit.

 

 

(7) \[\int_{a}^{b}{f(x)\,}dx=\int_{a}^{b}{f(a+b-x)dx}\].

 

 

(8) \[\int_{0}^{a}{\,\,x\,f(x)dx}=\frac{1}{2}a\int_{0}^{a}{\,\,f(x)dx}\], if \[f(a-x)=f(x)\].

 

 

(9) If \[f(x)\] is a periodic function with period T, then \[\int_{0}^{nT}{f(x)dx=n\int_{0}^{T}{\,\,f(x)dx}}\]

 

 

Deduction : If \[f(x)\] is a periodic function with period T, then

 

 

\[\int_{a}^{a+nT}{{}}f(x)\,dx=n\,\int_{0}^{T}{{}}f(x)\,dx\] , where \[n\,\in \,I\]

 

 

(a) If \[a=0,\] \[\int_{0}^{nT}{{}}f(x)\,dx=n\int_{0}^{T}{{}}f(x)\,dx,\] where \[n\in I\]

 

 

(b) If \[n=1,\] \[\int_{0}^{a+T}{{}}f(x)\,dx=\int_{0}^{T}{{}}f(x)\,dx\].

 

 

(10) \[\int_{mT}^{nT}{{}}f(x)\,dx=(n-m)\,\,\int_{0}^{T}{{}}f(x)\,dx,\] where \[n,\] \[m\in I\].

 

 

(11) \[\int_{a+nT}^{b+nT}{{}}f(x)\,dx=\int_{a}^{b}{{}}f(x)\,dx,\] where \[n\in \,I\].

 

 

(12) \[\int_{0}^{2k}{(x-[x])\,dx=k,}\] where \[k\] an integer, since \[x-[x]\] is a periodic function with period 1.

 

 

(13) If \[f(x)\] is a periodic function with period T, then \[\int_{a}^{a+T}{{}}f(x)\] is independent of a.

 

 

(14) \[\int_{a}^{b}{{}}f(x)\,dx=(b-a)\,\int_{0}^{1}{{}}f((b-a)\,x+a)\,dx\].


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