What is area y= 4/(sqrt(4-x^2)),(0,1) ?

The answer to area y= 4/(sqrt(4-x^2)),(0,1) is (2pi)/3

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area y= 4/(sqrt(4-x^2)),(0,1)

area

y = \frac{4}{4 - x ^{2}}, (0, 1)

\frac{2 π}{3}

Decimal Notation

2.09439 \dots

Solution steps

Apply the area formula:

\int_{0}^{1} \frac{4}{4 - x ^{2}} d x

Solve

\int_{0}^{1} \frac{4}{4 - x ^{2}} d x : \frac{2 π}{3}

The area is:

= \frac{2 π}{3}

t y^{^{'}} + y = 8

\frac{2 t}{t ^{2} + 4}

(5, - 3), (0, - 9)

\frac{\partial}{\partial y} (e^{3 x} \cdot tan (y))

\frac{d}{d x} ((x^{2} + 4)^{2} (2 x^{3} - 1)^{3})

What is area y= 4/(sqrt(4-x^2)),(0,1) ?
The answer to area y= 4/(sqrt(4-x^2)),(0,1) is (2pi)/3