_{1}

^{2}

_{-}

^{-}

_{c}

^{c}f(x) dx = ∫

_{1}

^{2}f(u-c) dx = 5

If ∫ f(x-c) dx from 1 to 2=5 where c is a constant, find ∫ f(x) dx from 1-c to 2-c.

Answer: 5

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The answer must be the same, 5, because you can use a u-substitution: u=x+c, du=dx.

It yields ∫_{1}^{2}_{-}^{-}_{c}^{c} f(x) dx = ∫_{1}^{2} f(u-c) dx = 5

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## Comments

To translate an equation's graph by <h,k> replace all the x's with x-h and all the y's with y-k.

In this case, the translation vector is <-c,0>, and

the function f(x-c) -> f((x--c)-c) = f(x),

the lower bound, x = 1 -> x--c = 1 or x = 1-c, and

the upper bound, x = 2 -> x--c = 2 or x = 2-c.

Since the function and the bounds are simply shifted horizontally on the coordinate plane, the integral value is the same, 5.