What is area x^{(4/3)},2x^{(1/3)},0,2 ?

The answer to area x^{(4/3)},2x^{(1/3)},0,2 is -(12*2^{1/3})/7+3*2^{1/3}

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area x^{(4/3)},2x^{(1/3)},0,2

area

x^{(\frac{4}{3})}, 2 x^{(\frac{1}{3})}, 0, 2

- \frac{12 \cdot 2 ^{\frac{1}{3}}}{7} + 3 \cdot 2^{\frac{1}{3}}

Decimal

1.61989 \dots

Solution steps

Apply the area formula:

\int_{0}^{2} x^{\frac{4}{3}} - 2 x^{\frac{1}{3}} d x

Solve

\int_{0}^{2} x^{\frac{4}{3}} - 2 x^{\frac{1}{3}} d x : - \frac{12 \cdot 2 ^{\frac{1}{3}}}{7} + 3 \cdot 2^{\frac{1}{3}}

The area is:

= - \frac{12 \cdot 2 ^{\frac{1}{3}}}{7} + 3 \cdot 2^{\frac{1}{3}}

(e^{y} + 1)^{2} e^{- y} d x + (e^{x} + 1)^{5} e^{- x} d y = 0

\int \frac{sec ( x )}{sec ( x ) - tan ( x )} d x

x \to 0 - lim (arcsin (e^{x}))

y = 6 x^{2} - 2 x, y = x^{3} - 6 x + 5

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

What is area x^{(4/3)},2x^{(1/3)},0,2 ?
The answer to area x^{(4/3)},2x^{(1/3)},0,2 is -(12*2^{1/3})/7+3*2^{1/3}