A) 4
B) 0
C) 1
D) 8
Correct Answer: A
Solution :
We have \[\int_{0}^{2\pi }{(\sin x+|\sin x|\,dx}\] \[=\int_{0}^{2\pi }{(\sin x+|\sin x|\,dx}+\int_{\pi }^{2\pi }{{}}\] \[(\sin x+|\sin x|)dx\] \[\left[ Using\,\int_{0}^{2a}{f(x)}\,dx=\int_{0}^{a}{f(x)\,dx+}\int_{a}^{2a}{f(x)\,dx} \right]\] \[=\int_{0}^{\pi }{(\sin x+\sin x)dx+\int_{\pi }^{2\pi }{(\sin x-\sin x)dx}}\] \[=\int_{0}^{\pi }{2\sin x\,dx+0}\] [\[|\sin x|\] is \[-ve\] in \[\pi \to 2\pi \] and \[(\sin x)\] is \[+ve\]in \[0\to \pi \]] \[=2\,\int_{0}^{\pi }{\sin x\,dx}\] \[=2\,[-\cos x]_{0}^{\pi }\] \[=-2\,(\cos \pi -\cos \,0)\] \[=-2\,(-1-1)\] \[=4\]
You need to login to perform this action.
You will be redirected in
3 sec
