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Question

Mathematics Question on applications of integrals

Area of the region bounded by two parabolas y=x2y\, =\, x^2 and x=y2x \,= \, y^2 is

A

13\frac{1}{3}

B

33

C

14\frac{1}{4}

D

44

Answer

13\frac{1}{3}

Explanation

Solution

The intersection point of parabolas y=x2y=x^{2} and x=y2x=y^{2} is y=(y2)2y=\left(y^{2}\right)^{2} y=y4\Rightarrow y=y^{4} y=0,y=1\Rightarrow y=0, y=1 x=0,x=1\Rightarrow x=0, x=1 So, the intersection point in O(0,0)O(0,0) and (1,1)(1,1). \therefore Required area =\int_\limits{0}^{1}\left(y_{2}-y_{1}\right) d x =\int_\limits{0}^{1}\left(\sqrt{x}-x^{2}\right) d x =[x3/23/2x33]01=[23x3/2x33]01=\left[\frac{x^{3 / 2}}{3 / 2}-\frac{x^{3}}{3}\right]_{0}^{1}=\left[\frac{2}{3} x^{3 / 2}-\frac{x^{3}}{3}\right]_{0}^{1} =[23(1)13(1)+00]=\left[\frac{2}{3}(1)-\frac{1}{3}(1)+0-0\right] =[2313]=13=\left[\frac{2}{3}-\frac{1}{3}\right]=\frac{1}{3} We know that, if parabolas are y2=4axy^{2}=4 a x and x2=4byx^{2}=4by, then area of bounded region is 4a4b3.\frac{4 a \cdot 4 b}{3}. \therefore Required area =1×13=13=\frac{1 \times 1}{3}=\frac{1}{3}