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.

  • 1
    $\begingroup$ Are $g, h$ in the interval $[a, b]$? $\endgroup$ Oct 19, 2014 at 20:40
  • $\begingroup$ Do you know anything about $g$ and $h$? If you still had $a$ and $b$ then the integral would be trivial, but without that I hardly see how this problem is doable. $\endgroup$ Oct 19, 2014 at 20:40
  • $\begingroup$ If $g, h \notin [a, b]$, it may be the case that $\int\limits_{g}^{h}f(t)\text{ d}t$ won't exist. $\endgroup$ Oct 19, 2014 at 20:52
  • $\begingroup$ Is $f$ periodic by any chance? $\endgroup$ Oct 19, 2014 at 20:58
  • $\begingroup$ We can take for example the function $f(x)=\frac{1}{x}$. It is continuous in $[1,2]$ and $\int_{a}^{b}f(u)du=ln(2)$. Now we can take $b-a=2-1=1-0=h-g$ and $\int_{g}^{h}f(t)dt$ is not defined... It seems that something is missing. $\endgroup$
    – Galc127
    Oct 19, 2014 at 21:01

1 Answer 1


$f$ 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: $$\int_{u = 2.0 - 1}^{u = 2.1 - 1} f(u) \frac{du}{2}$$


$\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.


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