What is area y=\sqrt[4]{x},y=sqrt(2-x),y=0 ?

The answer to area y=\sqrt[4]{x},y=sqrt(2-x),y=0 is 22/15

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area y=\sqrt[4]{x},y=sqrt(2-x),y=0

area

\frac{22}{15}

Decimal Notation

1.46666 \dots

Solution steps

To find the intersection points solve

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

2 - x = 0 : x = 2

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

\int_{1}^{2} 0 - 2 - x d x = \frac{2}{3}

= \frac{4}{5} + \frac{2}{3}

Simplify

= \frac{22}{15}

f (x) = \frac{2 x}{5 x - 2}

\frac{d}{d x} (\frac{1}{x + 5})

f (x) = 3 x^{4} - 12 x^{3} + 9

5 arcsin (x^{3})

x \to 0 lim (49 - x^{2})

What is area y=\sqrt[4]{x},y=sqrt(2-x),y=0 ?
The answer to area y=\sqrt[4]{x},y=sqrt(2-x),y=0 is 22/15