What is the integral from x to 1 of 6(1-x) ?

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

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integral from x to 1 of 6(1-x)

\int_{x}^{1} 6 (1 - x) d x

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

Solution steps

\int_{x}^{1} 6 (1 - x) d x

Take the constant out:

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

= 6 \cdot \int_{x}^{1} 1 - x 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

= 6 (\int_{x}^{1} 1 d x - \int_{x}^{1} x d x)

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

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

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

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