What is the integral from 1 to 6 of sqrt(3x-2) ?

The integral from 1 to 6 of sqrt(3x-2) is 14

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integral from 1 to 6 of sqrt(3x-2)

\int_{1}^{6} 3 x - 2 d x

14

Solution steps

\int_{1}^{6} 3 x - 2 d x

Apply u-substitution

: \int_{1}^{16} \frac{u}{3} d u

= \int_{1}^{16} \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_{1}^{16} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{16}

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{16}

Compute the boundaries:

42

= \frac{1}{3} \cdot 42

Simplify

= 14

\int_{1}^{2} \frac{7}{x} d x

\int_{0}^{2} \frac{x}{4 x + 1} d x

\int_{2}^{5} 2 x^{2} - x - 1 d x

\int_{0}^{1} 7 cos (x^{2}) d x

\int_{0}^{1} \frac{3 x ^{2}}{x ^{3} + 1} d x

What is the integral from 1 to 6 of sqrt(3x-2) ?
The integral from 1 to 6 of sqrt(3x-2) is 14