What is area y=1+\sqrt[3]{x},x=0,x=8,y=0 ?

The answer to area y=1+\sqrt[3]{x},x=0,x=8,y=0 is 20

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area y=1+\sqrt[3]{x},x=0,x=8,y=0

area

y = 1 + 3 x, x = 0, x = 8, y = 0

20

Solution steps

Apply the area formula:

\int_{0}^{8} 3 x + 1 d x

Solve

\int_{0}^{8} 3 x + 1 d x : 20

The area is:

= 20

\frac{d}{d x} (\frac{2 x ^{2} - 1}{( x - 1 ) ^{2}})

\frac{\partial}{\partial x} (x^{8} + 2^{y} + x^{y})

y^{^{'}} = \frac{x}{y ^{2}}

t \frac{d y}{d t} + 2 y = t^{2}

\int_{1}^{9} \frac{3 x - 2}{x} d x

What is area y=1+\sqrt[3]{x},x=0,x=8,y=0 ?
The answer to area y=1+\sqrt[3]{x},x=0,x=8,y=0 is 20