What is area y=((25-x^2)^{1/2}),y= 4/3 x,x=5 ?

The answer to area y=((25-x^2)^{1/2}),y= 4/3 x,x=5 is 50/3-(25pi)/4+25/2 arcsin(3/5)

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area y=((25-x^2)^{1/2}),y= 4/3 x,x=5

area

y = ((25 - x^{2})^{\frac{1}{2}}), y = \frac{4}{3} x, x = 5

\frac{50}{3} - \frac{25 π}{4} + \frac{25}{2} arcsin (\frac{3}{5})

Decimal

5.07547 \dots

Solution steps

Apply the area formula:

\int_{3}^{5} (25 - x^{2})^{\frac{1}{2}} - \frac{4}{3} x d x

Solve

\int_{3}^{5} (25 - x^{2})^{\frac{1}{2}} - \frac{4}{3} x d x : \frac{50}{3} - \frac{25 π}{4} + \frac{25}{2} arcsin (\frac{3}{5})

The area is:

= \frac{50}{3} - \frac{25 π}{4} + \frac{25}{2} arcsin (\frac{3}{5})

(\frac{1 + x ^{2}}{1 - x ^{2}})^{17}

\frac{d y}{d x} = sin (17 x)

x \to 1 lim (\frac{4 - x}{x - 1})

\int (x + 1) \cdot e^{4 x + 1} d x

3 4 x - 5

What is area y=((25-x^2)^{1/2}),y= 4/3 x,x=5 ?
The answer to area y=((25-x^2)^{1/2}),y= 4/3 x,x=5 is 50/3-(25pi)/4+25/2 arcsin(3/5)