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

The answer to area f(x)=((25-x^2)^{1/2}), 4/3 x,5 is 75/8-25/2 arcsin(3/5)

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

area

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

\frac{75}{8} - \frac{25}{2} arcsin (\frac{3}{5})

Decimal

1.33123 \dots

Solution steps

To find the intersection points solve

f_{i} (x) = f_{j} (x)

((25 - x^{2})^{\frac{1}{2}}) = \frac{4}{3} x : x = 3

((25 - x^{2})^{\frac{1}{2}}) = 5 : x = 0

\frac{4}{3} x = 5 : x = \frac{15}{4}

= \int_{0}^{3} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{3}^{\frac{15}{4}} ∣ f_{3} (x) - f_{2} (x) ∣ d x

= \int_{0}^{3} ((25 - x^{2})^{\frac{1}{2}}) - 5 d x + \int_{3}^{\frac{15}{4}} 5 - \frac{4}{3} x d x

\int_{0}^{3} ((25 - x^{2})^{\frac{1}{2}}) - 5 d x = - \frac{25}{2} arcsin (\frac{3}{5}) + 9

\int_{3}^{\frac{15}{4}} 5 - \frac{4}{3} x d x = \frac{3}{8}

= - \frac{25}{2} arcsin (\frac{3}{5}) + 9 + \frac{3}{8}

Simplify

= \frac{75}{8} - \frac{25}{2} arcsin (\frac{3}{5})

\frac{\partial}{\partial x} (a e^{2 x})

\int_{0}^{2} \frac{( x + 2 )}{x - 4} d x

\int \frac{1}{( a + b x ) ^{2}} d x

y = 5 - 9 x^{2}

\int \frac{1}{16 x ^{4} - 1} d x

What is area f(x)=((25-x^2)^{1/2}), 4/3 x,5 ?
The answer to area f(x)=((25-x^2)^{1/2}), 4/3 x,5 is 75/8-25/2 arcsin(3/5)