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

The integral from 2 to x of (2x-3)/6 is 1/6 (x^2-3x+2)

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

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

\frac{1}{6} (x^{2} - 3 x + 2)

Solution steps

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

\int_{2}^{x} 2 x d x = x^{2} - 4

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

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

Simplify

\frac{1}{6} (x^{2} - 4 - (3 x - 6)) : \frac{1}{6} (x^{2} - 3 x + 2)

= \frac{1}{6} (x^{2} - 3 x + 2)

\frac{d}{d x} (3 sec^{2} (x))

\frac{d y}{d x} = cos (5 x)

tan (u)

\frac{d}{d x} (ln (\frac{x - 1}{x}))

100 (1 + A e^{- t})^{- 1}

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