The parabolas y2=4x and x2=4y divide the square region bounded by the lines x=4, y=4, and the coordinate axes. If S1,S2,S3 are respectively the areas of these parts numbered from top to bottom, then S1:S2:S3 is:
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Official solution
Intersection points of x2=4y and y2=4x are \[\left( {0,0} \right)\] and \[\left( {4,4} \right).\] The graph is as shown in the figure.
By symmetry, we observe
\[{S_1} = {S_3} = \int\limits_0^4 {ydx = \int\limits_0^4 {{{{x^2}} \over 4}dx} } \]
\[ = \left[ {{{{x^3}} \over {12}}} \right]_0^4 = {{16} \over 3}\] sq. unit
Also \[{S_2} = \int\limits_0^4 {\left( {2\sqrt x - {{{x^2}} \over 4}} \right)} dx\]
\[\,\,\,\,\,\,\,\,\,\,\,\,\] \[ = \left[ {{{2{x^{{3 \over 2}}}} \over {{3 \over 2}}} - {{{x^3}} \over {12}}} \right]_0^4\]
\[\,\,\,\,\,\,\,\,\,\,\,\,\] \[ = {4 \over 3} \times 8 - {{16} \over 3} = {{16} \over 3}\]
\[\therefore\] S1:S2:S3=1:1:1