What is the integral from 0 to 12 of sqrt(3x) ?

The integral from 0 to 12 of sqrt(3x) is 48

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integral from 0 to 12 of sqrt(3x)

\int_{0}^{12} 3 x d x

48

Solution steps

\int_{0}^{12} 3 x d x

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \int_{0}^{12} 3 x d x

Take the constant out:

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

= 3 \cdot \int_{0}^{12} x d x

Apply the Power Rule

: [\frac{2}{3} x^{\frac{3}{2}}]_{0}^{12}

= 3 [\frac{2}{3} x^{\frac{3}{2}}]_{0}^{12}

Compute the boundaries:

163

= 3 \cdot 163

Simplify

= 48

f (t) = 2 cos (2 t)

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

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

x \to 2 lim (f (x) + 5 g (x))

What is the integral from 0 to 12 of sqrt(3x) ?
The integral from 0 to 12 of sqrt(3x) is 48