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

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

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

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

0

Solution steps

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

Take the constant out:

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

= f \cdot \int_{0}^{2} - 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

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

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

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

= f (- 2 + 2)

Simplify

= 0

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