What is area y= 4/3 ,y= 1/(sqrt(9-x^2)) ?

The answer to area y= 4/3 ,y= 1/(sqrt(9-x^2)) is 2sqrt(15)-2arcsin((sqrt(15))/4)

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area y= 4/3 ,y= 1/(sqrt(9-x^2))

area

y = \frac{4}{3}, y = \frac{1}{9 - x ^{2}}

215 - 2 arcsin (\frac{15}{4})

Decimal Notation

5.10973 \dots

Solution steps

Apply the area formula:

\int_{- \frac{3 15}{4}}^{\frac{3 15}{4}} \frac{4}{3} - \frac{1}{9 - x ^{2}} d x

Solve

\int_{- \frac{3 15}{4}}^{\frac{3 15}{4}} \frac{4}{3} - \frac{1}{9 - x ^{2}} d x : 215 - 2 arcsin (\frac{15}{4})

The area is:

= 215 - 2 arcsin (\frac{15}{4})

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

y^{^{'}} = 1 - x + 4 y

\frac{\partial}{\partial x} (5 e^{x^{6} y})

\frac{3 x + 6}{2 x + 3}

\int_{- 4}^{2} - \frac{y ^{2}}{2} - y + 4 d y

What is area y= 4/3 ,y= 1/(sqrt(9-x^2)) ?
The answer to area y= 4/3 ,y= 1/(sqrt(9-x^2)) is 2sqrt(15)-2arcsin((sqrt(15))/4)