I'm given that a function $f$ is continuous in $[a, b]$ and given a value $\int_a^b f(u)du = c$. Then I'm asked to calculate $\int_g^h f(t)dt$. I'm looking at the fundamental theorem of calculus but I can't get a clue about how to proceed.
I did one exercise that was easier, the integral to calculate was of $f(x -2)$ and the given integral was of $f(x)$. A simple substitution by $u$ solved it. But now I have a different case, the given integral has the variable $u$ and the integral to be calculated has $x$.
Edit: in both exercises, g and h weren't located in between [a, b]. The only pattern I could notice is that $b – a = h – g$. So the limits of integration were simply shifted to the right or to the left.
Hmm… don't know from where this exercise comes from. When I did the first exercise, I asked "does it matter if $f$ is odd or even?". I made a guess and saw that, whatever the function was, x – 2 shifted the whole graph by two units and since the limits of integration were also shifted by two, I could convince myself that the value of the integral was kept the same with the shift and the substitution.
Best Answer
$f$ is continuous in $[-1, 1]$. Calculate $\int_0^1 f(2x -1) dx$, given that $\int_{-1}^1 f(u) du = 5$
$u = 2x - 1 \Rightarrow x = \frac{u+1}{2}$
$du = 2dx$
This is where I was confused: $$\int_{u = 2.0 - 1}^{u = 2.1 - 1} f(u) \frac{du}{2}$$
So:
$\int_{-1}^1 f(u) \frac{du}{2} $
$\frac{1}{2}\int_{-1}^1 f(u) du = \frac{5}{2}$
I guess that if I face another problem like this, but a substitution falls in an interval, say [-2, 3] whereas the given definite integral is from 0 to 2, the exercise must contain a typo.