The area included between the parabola y= x^24a and the curve y= 8a^3(x^2+4a^2) is

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Correct answer: D. (d) y= x^24a........(i) y= 8a^3x^2+4a^2......(ii) For point of intersection equation (i) and (ii), x^24a= 8a^3x^2+4a^2 ⇒x^4+4a^2x^2-32a^2=0 ⇒(x^2-4a^2)(x^2+8a^2)=0 ⇒x^2-4a^2=0 or x^2+8a^2=0 ⇒x^2=4a^2 (x^2=-8a^2 is not possible ) ⇒x= 2a So, we take limits from 0 to 2a. Now, Area enclosed by the two curves A=2 0.05em 0.05em( 8a^3(x^2+4a^2)- x^24a) dx=2( 0.05em 0.05em 8a^3(x^2+4a^2) dx- 0.05em 0.05em x^24a dx) =2(8a^3 _0^2a 0.05em 1x^2+ (2a^2 ) dx- 14a _0^2a 0.05emx^2 dx)=2(8a^3 ( 12a tan^-1 ⁡ ( x2a ) )_0^2a- 14a ( x^33 )_0^2a)=2 ( 8a^32a( tan^-1 ⁡ ( 2a2a )- tan^-1 ⁡ ( 02a )) ) ( 14a( (2a )^33- (0 )^33) ) =2(4a^2( tan^-1 ⁡ (1 )- tan^-1 ⁡ (0 ))- 14a( 8a^33-0))=2(4a^2( 4-0)- 14a ( 8a^33 ))=2 (a^2( - 23) )

$The area included between the parabola y = \frac{x ^{2}}{4 a} and the curve y = \frac{8 a ^{3}}{( x ^{2} + 4 a ^{2} )} is$

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