Let's rewrite the function x = 2 y . x=2\sqrt{y}. x = 2 y .
x 2 = ( 2 y ) 2 , x ≥ 0 , y ≥ 0 ; x^2=(2\sqrt{y})^2, x\geq 0, y\geq 0; x 2 = ( 2 y ) 2 , x ≥ 0 , y ≥ 0 ;
x 2 = 4 y , x ≥ 0 , y ≥ 0 ; x^2=4y, x\geq 0, y\geq 0; x 2 = 4 y , x ≥ 0 , y ≥ 0 ;
y = 1 4 x 2 , x ≥ 0 , y ≥ 0. y=\frac{1}{4}x^2, x\geq 0, y\geq 0. y = 4 1 x 2 , x ≥ 0 , y ≥ 0.
Let's find the interseptions of the curves:
1 4 x 2 = 2 x , \frac{1}{4}x^2=2\sqrt{x}, 4 1 x 2 = 2 x ,
( 1 4 x 2 ) 2 = ( 2 x ) 2 , (\frac{1}{4}x^2)^2=(2\sqrt{x})^2, ( 4 1 x 2 ) 2 = ( 2 x ) 2 ,
1 16 x 4 = 4 x , \frac{1}{16}x^4=4x, 16 1 x 4 = 4 x ,
x 4 = 64 x , x^4=64x, x 4 = 64 x ,
x 4 − 64 x = 0 , x^4-64x=0, x 4 − 64 x = 0 ,
x ( x 3 − 64 ) = 0 , x(x^3-64)=0, x ( x 3 − 64 ) = 0 ,
x = 0 x=0 x = 0 or x 3 − 64 = 0 , x^3-64=0, x 3 − 64 = 0 ,
x = 0 x=0 x = 0 or x = 4. x=4. x = 4.
A = ∫ 0 4 2 x d x − ∫ 0 4 1 4 x 2 d x = A=\int_0^42\sqrt{x}dx-\int_0^4 \frac{1}{4}x^2dx= A = ∫ 0 4 2 x d x − ∫ 0 4 4 1 x 2 d x =
= 2 ∫ 0 4 x d x − 1 4 ∫ 0 4 x 2 d x = =2\int_0^4\sqrt{x}dx- \frac{1}{4}\int_0^4x^2dx= = 2 ∫ 0 4 x d x − 4 1 ∫ 0 4 x 2 d x =
= 2 ⋅ 2 3 x x ∣ 0 4 − 1 4 x 3 3 ∣ 0 4 = =2\cdot \frac{2}{3}x\sqrt{x} |_0^4-\frac{1}{4}\frac{x^3}{3}|_0^4= = 2 ⋅ 3 2 x x ∣ 0 4 − 4 1 3 x 3 ∣ 0 4 =
= 4 3 ⋅ 4 4 − 1 4 ⋅ 4 3 3 ∣ 0 4 = 32 3 − 16 3 = 16 3 = 5 1 3 . =\frac{4}{3}\cdot 4\sqrt{4}-\frac{1}{4}\cdot \frac{4^3}{3}|_0^4=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}=5\frac{1}{3}. = 3 4 ⋅ 4 4 − 4 1 ⋅ 3 4 3 ∣ 0 4 = 3 32 − 3 16 = 3 16 = 5 3 1 .
Answer : 5 1 3 . 5\frac{1}{3}. 5 3 1 .
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