What is area y= 9/(sqrt(9x-x^2)),x=1,x=7.5 ?

The answer to area y= 9/(sqrt(9x-x^2)),x=1,x=7.5 is 9(arcsin(2/3)+arcsin(7/9))

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area y= 9/(sqrt(9x-x^2)),x=1,x=7.5

area

y = \frac{9}{9 x - x ^{2}}, x = 1, x = 7.5

9 (arcsin (\frac{2}{3}) + arcsin (\frac{7}{9}))

Solution steps

Apply the area formula:

\int_{1}^{7.5} \frac{9}{9 x - x ^{2}} d x

Solve

\int_{1}^{7.5} \frac{9}{9 x - x ^{2}} d x : 9 (arcsin (\frac{2}{3}) + arcsin (\frac{7}{9}))

The area is:

= 9 (arcsin (\frac{2}{3}) + arcsin (\frac{7}{9}))

\int_{- 1}^{1} 2 - x^{2} d x

\int_{- 1}^{2} x^{2} - 2 x d x

(1 + x)^{n} 0

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

\frac{\partial}{\partial x} (6 y + 1)

What is area y= 9/(sqrt(9x-x^2)),x=1,x=7.5 ?
The answer to area y= 9/(sqrt(9x-x^2)),x=1,x=7.5 is 9(arcsin(2/3)+arcsin(7/9))