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

  • question_answer Area under the curve \[y=\sin 2x+\cos 2x\] between \[x=0\] and \[x=\frac{\pi }{4},\] is                                 [AI CBSE 1979]

    A)            2 sq. unit                                

    B)            1 sq. unit

    C)            3 sq. unit                                

    D)            4 sq. unit

    Correct Answer: B

    Solution :

                       Required area \[=\int_{0}^{\pi /4}{(\sin 2x+\cos 2x)dx}\]                                \[=\left[ -\frac{\cos 2x}{2}+\frac{\sin 2x}{2} \right]_{0}^{\pi /4}\]                                \[=\frac{1}{2}\left[ -\cos \frac{\pi }{2}+\sin \frac{\pi }{2}+\cos 0-\sin 0 \right]=1\,sq.\]unit.


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