What is area y=(x^3)/9 ,0<= x<= 2 ?

The answer to area y=(x^3)/9 ,0<= x<= 2 is 4/9

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area y=(x^3)/9 ,0<= x<= 2

area

y = \frac{x ^{3}}{9}, 0 \leq x \leq 2

\frac{4}{9}

Decimal Notation

0.44444 \dots

Solution steps

Apply the area formula:

\int_{0}^{2} \frac{1}{9} x^{3} d x

Solve

\int_{0}^{2} \frac{1}{9} x^{3} d x : \frac{4}{9}

The area is:

= \frac{4}{9}

t y^{^{'}} + 7 y = t^{2} - t + 1, y (1) = \frac{1}{7}

\int tan^{2} (d) x d x

\frac{d}{d x} (ln (x))

x \to 2 - lim (- \frac{1}{x ^{2} - 4})

f (x) = 3 x - \frac{1}{x}

What is area y=(x^3)/9 ,0<= x<= 2 ?
The answer to area y=(x^3)/9 ,0<= x<= 2 is 4/9