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

The integral from 2 to 6 of 1/(sqrt(3x-2)) is 4/3

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

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

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

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

Apply u-substitution

: \int_{4}^{16} \frac{1}{3 u} d u

= \int_{4}^{16} \frac{1}{3 u} 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_{4}^{16} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{4}^{16}

= \frac{1}{3} [2 u^{\frac{1}{2}}]_{4}^{16}

Apply radical rule:

a^{\frac{1}{n}} = n a

= \frac{1}{3} [2 u]_{4}^{16}

Compute the boundaries:

4

= \frac{1}{3} \cdot 4

Simplify

= \frac{4}{3}

\frac{d}{d t} (2 cos (t) + 2 sin (t))

F (x) = (9 x^{6} + 4 x^{3})^{4}

y^{^{''}} - 2 y^{^{'}} - 3 y = 16 x^{2} e^{- x}

y^{^{''}} + 4 y^{^{'}} + 5 y = e^{x} + 5 x

\frac{1}{( s ^{2} + 4 ) \cdot s ^{2} \cdot e}

What is the integral from 2 to 6 of 1/(sqrt(3x-2)) ?
The integral from 2 to 6 of 1/(sqrt(3x-2)) is 4/3