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 Notation

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}

1 - 9 x^{2} y^{^{'}} = x

\frac{d}{d x} (\frac{∣ x ∣}{2 - x ^{2}})

\frac{d y}{d t} + 5 y = 5

3 (1 + t^{2}) (\frac{d y}{d t}) = 2 t y (y^{3} - 1)

y^{^{''}} + 81 y = - 81 sec^{2} (9 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