question_answer

The figure shows as triangle AOB and the parabola\[y={{x}^{2}}\]. The ratio of the area of the triangle AOB to the area of the region AOB of the parabola \[y={{x}^{2}}\] is equal to

A) \[\frac{3}{5}\]

B) \[\frac{3}{4}\]

C) \[\frac{7}{8}\]

D) \[\frac{5}{6}\]

Correct Answer: B

Solution :

[b] Area of \[\Delta AOB=\frac{1}{2}\times 2a\times {{a}^{2}}={{a}^{3}}\] units Area of region AOB \[=2\int\limits_{0}^{{{a}^{2}}}{xdy=2\int\limits_{0}^{{{a}^{2}}}{\sqrt{y}\,dy=2\left[ \frac{{{y}^{3/2}}}{3/2} \right]_{0}^{{{a}^{2}}}=\frac{4}{3}{{a}^{3}}}}\] units \[\therefore \] ratio of areas \[=\frac{{{a}^{3}}}{\frac{4}{3}{{a}^{3}}}=\frac{3}{4}\]