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

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

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

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

x^{2} - 2 x + 1

Solution steps

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

Take the constant out:

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

= 2 \cdot \int_{1}^{x} x - 1 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

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

\int_{1}^{x} x d x = \frac{x ^{2} - 1}{2}

\int_{1}^{x} 1 d x = x - 1

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

Simplify

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

= x^{2} - 2 x + 1

f (x) = arctan (x^{10})

3 t^{4} + 7 t^{2} - 5

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

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