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 1/2/2014 | Sun from Los Angeles, CA | 1 Answer | 0 Votes Mark favorite Subscribe Comment
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 1/2/2014 | Roman C. Comment Comments Thanks. 1/2/2014 | Sun from Los Angeles, CA Another perspective: 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. 1/5/2014 | Steve S. Comment
Another perspective: 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. 1/5/2014 | Steve S.
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.