What is the integral of sqrt(16x^5) ?

The integral of sqrt(16x^5) is 8/7 x^{7/2}+C

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integral of sqrt(16x^5)

\int 16 x^{5} d x

\frac{8}{7} x^{\frac{7}{2}} + C

Solution steps

\int 16 x^{5} d x

16 x^{5} = 16 x^{5}

= \int 16 x^{5} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 16 \cdot \int x^{5} d x

16 = 4

= 4 \cdot \int x^{5} d x

x^{5} = x^{\frac{5}{2}},

assuming

x \geq 0

= 4 \cdot \int x^{\frac{5}{2}} d x

Apply the Power Rule

= 4 \cdot \frac{2}{7} x^{\frac{7}{2}}

Simplify

4 \cdot \frac{2}{7} x^{\frac{7}{2}} : \frac{8}{7} x^{\frac{7}{2}}

= \frac{8}{7} x^{\frac{7}{2}}

Add a constant to the solution

= \frac{8}{7} x^{\frac{7}{2}} + C

What is the integral of sqrt(16x^5) ?
The integral of sqrt(16x^5) is 8/7 x^{7/2}+C