What is the integral from 0 to x of 6(1-y) ?

The integral from 0 to x of 6(1-y) is 6(x-(x^2)/2)

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integral from 0 to x of 6(1-y)

\int_{0}^{x} 6 (1 - y) d y

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

Solution steps

\int_{0}^{x} 6 (1 - y) d y

Take the constant out:

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

= 6 \cdot \int_{0}^{x} 1 - y d y

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_{0}^{x} 1 d y - \int_{0}^{x} y d y)

\int_{0}^{x} 1 d y = x

\int_{0}^{x} y d y = \frac{x ^{2}}{2}

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

\int_{0}^{2} x (x - 2) d x

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\int_{- 9}^{9} ∣ x ∣ d x

What is the integral from 0 to x of 6(1-y) ?
The integral from 0 to x of 6(1-y) is 6(x-(x^2)/2)