A) \[{{I}_{1}}+{{I}_{2}}=\frac{\pi }{2}\]
B) \[{{I}_{2}}-{{I}_{1}}=\frac{\pi }{2}\]
C) \[{{I}_{1}}+{{I}_{2}}=0\]
D) \[{{I}_{1}}={{I}_{2}}\]
Correct Answer: B
Solution :
Since, \[{{I}_{1}}=\int_{0}^{\pi /2}{x\,\sin \,x\,dx}\] and \[{{I}_{2}}=\int_{0}^{\pi /2}{x\,\cos \,x\,dx}\] \[\therefore \] \[{{I}_{1}}=\int_{0}^{\pi /2}{x\,sin\,x\,dx}\] \[\Rightarrow \] \[{{I}_{1}}=-[x\,\cos \,x]_{0}^{\pi /2}-\int_{0}^{\pi /2}{-\cos \,x\,dx}\] \[=[0-0]+[\sin \,x]_{0}^{\pi /2}\] \[=1\] ??..(i) and \[{{I}_{2}}=\int_{0}^{\pi /2}{x\,\,\cos \,\,x\,\,dx}\] \[=[x\,\sin \,x]_{0}^{\pi /2}\int_{0}^{\pi /2}{\cos \,\,x\,\,dx}\] \[=\left( \frac{x}{2}-0 \right)+[\sin x]_{0}^{\pi /2}\] \[=\frac{\pi }{2}+1\] From (i) and (ii) \[{{I}_{2}}-{{I}_{1}}=\frac{\pi }{2}\]
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