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

  • question_answer The area formed by triangular shaped region bounded by the curves \[y=\sin x,\,y=\cos x\] and \[x=0\] is           [MP PET 2000]

    A)            \[x={{y}^{2}}\]                    

    B)            1

    C)            \[\sqrt{2}\]                           

    D)            \[1+\sqrt{2}\]

    Correct Answer: A

    Solution :

               Given required area has been shown in the figure.            \[x=\frac{\pi }{4}\]  is the point of intersection of both curve            \[\therefore \]Required area = \[\int_{0}^{\pi /4}{(\cos x-\sin x)\,dx}\]                                      \[=[\sin x+\cos x]_{0}^{\pi /4}\]\[=\left[ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1 \right]\]                                      =\[\frac{2}{\sqrt{2}}-1=\sqrt{2}-1\].

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