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

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

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integral from 3 to 4 of (2x-3)/6

\int_{3}^{4} \frac{2 x - 3}{6} d x

\frac{2}{3}

Decimal

0.66666 \dots

Solution steps

\int_{3}^{4} \frac{2 x - 3}{6} d x

Take the constant out:

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

= \frac{1}{6} \cdot \int_{3}^{4} 2 x - 3 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{6} (\int_{3}^{4} 2 x d x - \int_{3}^{4} 3 d x)

\int_{3}^{4} 2 x d x = 7

\int_{3}^{4} 3 d x = 3

= \frac{1}{6} (7 - 3)

Simplify

\frac{1}{6} (7 - 3) : \frac{2}{3}

= \frac{2}{3}

\int_{2}^{3} (3 - x) d x

\frac{d y}{d x} = \frac{x y + 2}{x + 1}

\frac{d}{d t} (t - cos (t))

x \to + 2 lim (x 13 - x^{2})

\int 1 + \frac{2}{x} d x

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