What is the integral from 1 to 4 of sqrt(3x-3) ?

The integral from 1 to 4 of sqrt(3x-3) is 6

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integral from 1 to 4 of sqrt(3x-3)

\int_{1}^{4} 3 x - 3 d x

6

Solution steps

\int_{1}^{4} 3 x - 3 d x

Apply u-substitution

: \int_{0}^{9} \frac{u}{3} d u

= \int_{0}^{9} \frac{u}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{0}^{9} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9}

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9}

Compute the boundaries:

18

= \frac{1}{3} \cdot 18

Simplify

= 6

\int_{- \infty}^{- 1} \frac{e ^{x}}{x} d x

\int_{0}^{6} \frac{x}{18} d x

\int_{- 0.5}^{0.5} 1 - x^{2} d x

\int_{1}^{3 x} ln (x) d x

\int_{- 1}^{4} x + 4 - x^{2} + 2 x d x

What is the integral from 1 to 4 of sqrt(3x-3) ?
The integral from 1 to 4 of sqrt(3x-3) is 6