Suppose I have an integral to evaluate, which is of the form ,

$$\int_0^x xf(t)\,dt$$

So , is it correct to consider the $x$ in the integrand as a constant and remove it outside the integral? I think the answer is yes because it does not change with $t$ . But I’m still confused as the upper limit also contains $x$ . Please help me with this .

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    After toying around on Desmos for a bit I'm convinced that the x in the integrand can be considered a constant and be removed. Since it's dt and not dx this makes sense. Not sure how to show this rigorously however. – Bryden C Aug 11 at 1:57
  • @Bryden thanks for your help :) – Aditi Aug 11 at 2:02
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    Is $x$ a non-constant function of $t$, or is $x$ not a function of $t$? – Chase Ryan Taylor Aug 11 at 2:19
  • @Chase Ryan usually we are given $g(x)$ to be given equal to that integral and we find the monotonicity of $g$ I think $x$ and $t$ are independent – Aditi Aug 11 at 2:33
up vote 3 down vote accepted

If you just want to calculate the value of the integral, this should be fine if I'm not mistaken, as you might also call the upper bound $u$ and later set $u=x$ again after calculation. More precisely, to evaluate the integral, I'm parametrizing the upper bound to $u$ s.t. we consider

$$h(u)=\int_0^uxf(t)dt=x\int_0^uf(t)dt$$

Thus, you may evaluate $\int_0^uf(t)dt$ in dependence on $u$ and then derive the value of the original integral via $h(x)$.

However, if you are working with the by the integral induced function

$$g(x)=\int_0^xxf(t)dt$$

and want to perform more complex tasks like differentiation or integration of $g$, then this is called a parametric integral and the answer is no. But in this case, $x$ would also less be considered a constant value than a variable(for this respective function).

  • This really makes sense .Usually we consider the parametric form because we have to find the monotonicity of $g(x)$ . So we need to differentiate it using Leibniz Rule. Thanks for the help ! – Aditi Aug 11 at 2:01
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    You're very welcome. I've just edited the answer to include some more information. Note, that this reparametrization works using $u$ works, as you're just interested in the value at the point $x$ for $h$. $h$ is of course only sparsely related to your integral and thus can not be used for things like differentiability. – zzuussee Aug 11 at 2:06
  • Absolutely . Thanks again. – Aditi Aug 11 at 2:09

...[I]s it correct to consider the $x$ in the integrand as a constant and remove it outside the integral?

Well, the answer depends on how you regard $x,$ depending on the problem you want to solve. Since you want to evaluate the integral, it would seem that $x$ should be regarded as some (as yet unspecified, but fixed) number. In this case you then proceed as usual. It is then clear that for each $x$ (regarded as a number) for which the integral is defined, there is no difference in the evaluation.

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    Okay thank you for your help ! – Aditi Aug 11 at 5:34

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