What is the integral from 2 to x of 1/8 (x-2) ?

The integral from 2 to x of 1/8 (x-2) is 1/16 (x^2-4x+4)

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integral from 2 to x of 1/8 (x-2)

\int_{2}^{x} \frac{1}{8} (x - 2) d x

\frac{1}{16} (x^{2} - 4 x + 4)

Solution steps

\int_{2}^{x} \frac{1}{8} (x - 2) d x

Take the constant out:

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

= \frac{1}{8} \cdot \int_{2}^{x} x - 2 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}{8} (\int_{2}^{x} x d x - \int_{2}^{x} 2 d x)

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

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

= \frac{1}{8} (\frac{x ^{2}}{2} - 2 - (2 x - 4))

Simplify

\frac{1}{8} (\frac{x ^{2}}{2} - 2 - (2 x - 4)) : \frac{1}{16} (x^{2} - 4 x + 4)

= \frac{1}{16} (x^{2} - 4 x + 4)

n = 1 \sum \infty (- 6)^{n} x^{n}

\int \frac{x ^{2}}{1 - 2 x ^{3}} d x

\frac{d}{d x} (\frac{e ^{x}}{7 x ^{2} + 9})

x \to - 7 lim (12)

x^{2} - 8 x + 17

What is the integral from 2 to x of 1/8 (x-2) ?
The integral from 2 to x of 1/8 (x-2) is 1/16 (x^2-4x+4)