What is the integral from 0 to 39 of 0.31sqrt(x) ?

The integral from 0 to 39 of 0.31sqrt(x) is 50.33468…

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integral from 0 to 39 of 0.31sqrt(x)

\int_{0}^{39} 0.31 x d x

50.33468 \dots

Solution steps

\int_{0}^{39} 0.31 x d x

Take the constant out:

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

= 0.31 \cdot \int_{0}^{39} x d x

Apply the Power Rule

: [\frac{2}{3} x^{\frac{3}{2}}]_{0}^{39}

= 0.31 [\frac{2}{3} x^{\frac{3}{2}}]_{0}^{39}

Simplify

= 0.31 [0.66666 \dots x^{\frac{3}{2}}]_{0}^{39}

Compute the boundaries:

162.36994 \dots

= 0.31 \cdot 162.36994 \dots

Simplify

= 50.33468 \dots

n = 1 \sum \infty n (- 1)^{n}

sin^{3} (t)

\int \frac{1}{8 y} d y

\int sec^{2} (x) tan^{5} (x) d x

\frac{d y}{d x} + 2 x y = - x

What is the integral from 0 to 39 of 0.31sqrt(x) ?
The integral from 0 to 39 of 0.31sqrt(x) is 50.33468…