What is area x=y^{1/2},x=2[0,2] ?

The answer to area x=y^{1/2},x=2[0,2] is 0+(((2[0,2])^3)/3-0)

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area x=y^{1/2},x=2[0,2]

area

x = y^{\frac{1}{2}}, x = 2 [0, 2]

0 + (\frac{( 2 [ 0 , 2 ] ) ^{3}}{3} - 0)

Solution steps

Apply the area formula:

\int_{0}^{2 [0, 2]} ∣ x ∣^{2} d x

Solve

\int_{0}^{2 [0, 2]} ∣ x ∣^{2} d x : 0 + (\frac{( 2 [ 0 , 2 ] ) ^{3}}{3} - 0)

The area is:

= 0 + (\frac{( 2 [ 0 , 2 ] ) ^{3}}{3} - 0)

f (x) = x^{2}, at x = 0.2

f (x) = \frac{7}{x ^{2}}

\int \frac{1}{3} sec (x) tan (x) d x

\frac{d}{d x} (- \frac{1}{x ^{2}} + 1)

x \to \infty lim (1 - 1^{x})

What is area x=y^{1/2},x=2[0,2] ?
The answer to area x=y^{1/2},x=2[0,2] is 0+(((2[0,2])^3)/3-0)