What is area y=2x^{1/3},0<= x<= 32 ?

The answer to area y=2x^{1/3},0<= x<= 32 is 96*2^{2/3}

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area y=2x^{1/3},0<= x<= 32

area

y = 2 x^{\frac{1}{3}}, 0 \leq x \leq 32

96 \cdot 2^{\frac{2}{3}}

Decimal

152.39050 \dots

Solution steps

Apply the area formula:

\int_{0}^{32} 2 x^{\frac{1}{3}} d x

Solve

\int_{0}^{32} 2 x^{\frac{1}{3}} d x : 96 \cdot 2^{\frac{2}{3}}

The area is:

= 96 \cdot 2^{\frac{2}{3}}

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What is area y=2x^{1/3},0<= x<= 32 ?
The answer to area y=2x^{1/3},0<= x<= 32 is 96*2^{2/3}