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    Topics || Problems

    What is the area bounded by the curve \(y = \sin {x}\) and \(y = \cos {x}\) and the x-axis in the first quadrant.
    Solution
    y =sin x and y =cos x

    The area will be the shaded part. To calculate, divide into two parts, from (1) \(x=0\) to \(x=\frac{\pi}{4}\) and from (2) \(x=\frac{\pi}{4}\) to \(x=\frac{\pi}{2}\)

    y =sin x and y =cos x

    The first area is the area bounded by \(y = \sin x \) and the \(x -axis\) from \(x=0\) to \(x=\frac{\pi}{4}\) .

    y =sin x and y =cos x

    The second area is the area bounded by \(y = \cos x \) and the \(x -axis\) from \(x= \frac{\pi}{4}\) to \(x=\frac{\pi}{2}\).

    y =sin x and y =cos x

    \(A_1 = \int_{0}^{\frac{\pi}{4}} { \sin{x} ~dx} \)

    \(A_1 =\cos {x} |_{0}^{\frac{\pi}{4}} \)

    \(A_1 =| (\cos{\frac{\pi}{4}}) - (\cos {0} )|\)

    \(A_1 = 0.293\)


    \(A_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} { \cos{x} ~dx} \)

    \(A_2 = - \sin {x} |_{\frac{\pi}{4}}^{\frac{\pi}{2}} \)

    \(A_2 =|- (\sin {\frac{\pi}{2}} ) + (\sin {\frac{\pi}{4}} )|\)

    \(A_2 = 0.293\)


    Answer: \(A = 0.293 +0.293 = 0.586~\text{sq area}\)