What is the integral from 0 to 1 of 2(1-x) ?

The integral from 0 to 1 of 2(1-x) is 1

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integral from 0 to 1 of 2(1-x)

\int_{0}^{1} 2 (1 - x) d x

1

Solution steps

\int_{0}^{1} 2 (1 - x) d x

Take the constant out:

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

= 2 \cdot \int_{0}^{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

= 2 (\int_{0}^{1} 1 d x - \int_{0}^{1} x d x)

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

\int_{0}^{1} x d x = \frac{1}{2}

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

Simplify

2 (1 - \frac{1}{2}) : 1

= 1

\int \frac{4 x}{4 x ^{2} + 1} d x

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

y = e^{- x} cos (2 x)

\frac{1}{s ^{4} + s ^{2}}

\int (x^{2} + 2 x) d x

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