What is the integral of \sqrt[3]{r^2} ?

The integral of \sqrt[3]{r^2} is 3/5 r^{5/3}+C

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integral of \sqrt[3]{r^2}

\int 3 r^{2} d r

\frac{3}{5} r^{\frac{5}{3}} + C

Solution steps

\int 3 r^{2} d r

3 r^{2} = r^{\frac{2}{3}},

assuming

r \geq 0

= \int r^{\frac{2}{3}} d r

Apply the Power Rule

: \frac{3}{5} r^{\frac{5}{3}}

= \frac{3}{5} r^{\frac{5}{3}}

Add a constant to the solution

= \frac{3}{5} r^{\frac{5}{3}} + C

f (x) = x^{2} + 5, at x = 2

y = x^{3} - 4 x, (2, 0)

y^{^{''}} + 25 y = - 30 sin (5 t)

\int_{2}^{4} \frac{1}{( x - 1 ) ( x - 4 )} d x

\frac{d}{d x} (2 (1 + cos (x)))

What is the integral of \sqrt[3]{r^2} ?
The integral of \sqrt[3]{r^2} is 3/5 r^{5/3}+C