What is area y=x^{4/3},y=2x^{1/3} ?

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

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area y=x^{4/3},y=2x^{1/3}

area

y = x^{\frac{4}{3}}, y = 2 x^{\frac{1}{3}}

- \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}}

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