JEE Main & Advanced Mathematics Definite Integration Question Bank Area Bounded by Region, Volume and Surface Area of Solids of Revolution

  • question_answer The area bounded by curves \[y=\cos x\] and \[y=\sin x\] and ordinates \[x=0\] and \[x=\frac{\pi }{4}\] is                               [Karnataka CET 2002]

    A)            \[\sqrt{2}\]                           

    B)            \[\sqrt{2}+1\]

    C)            \[\sqrt{2}-1\]                       

    D)            \[\sqrt{2}(\sqrt{2}-1)\]

    Correct Answer: C

    Solution :

               Given equations of curves \[y=\cos x\] and \[y=\sin x\] and ordinates \[x=0\] to \[x=\frac{\pi }{4}.\]We know that area bounded by the curves \[=\int_{{{x}_{1}}}^{{{x}_{2}}}{ydx=\int_{0}^{\pi /4}{\cos xdx-\int_{0}^{\pi /4}{\,\,\,\,\,\,\sin x\,dx}}}\]                    \[=[\sin x]_{0}^{\pi /4}-[-\cos x]_{0}^{\pi /4}\]                    \[=\left( \sin \frac{\pi }{4}-\sin 0 \right)+\left( \cos \frac{\pi }{4}-\cos 0 \right)=\left( \frac{1}{\sqrt{2}}-0 \right)+\left( \frac{1}{\sqrt{2}}-1 \right)\]                    \[=\sqrt{2}-1\].


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