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

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

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

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

2 k

Solution steps

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

Take the constant out:

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

= k \cdot \int_{- 1}^{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

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

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

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

= k (2 - 0)

Simplify

= 2 k

\int_{0}^{1} (x^{3} + 4 x) d x

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\int_{1}^{2} 3 x - x^{2} + 2 d x

\int_{0}^{1} 2 + \frac{1}{4 x} d x

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