# Change of variables for an integral

I have an integral where I need to change variables. The integral has the form,

$$\int_0^x f(x,t) dt$$ .

I change variables/rescale by setting $$\tilde{t}=xt$$, which means $$d\tilde{t}=xdt$$. Would the new integral have the following form & bounds:

$$\int_0^{x^2} f(x,\tilde{t}) d\tilde{t}$$ ?

• Why did you substitute $t$ with $\tilde{t}$ directly, clearly $t = \tilde{t}/x$. Feb 8 '19 at 1:25
• You have $\int_0^x f(x, t)dt$ and you want to make the change of variable $s= xt$. Yes, $ds= x dt$ so that $dt= \frac{1}{x}ds$. When t= 0, s= 0 and wen t= x, $s= x^2$. The integral is $\frac{1}{x}\int_0^{x^2} f(x, s/x)ds$. Feb 8 '19 at 1:26

As you correctly computed, $$d\widetilde{t}=xdt$$, so $$dt=d\widetilde{t}/x$$. We can see by $$\widetilde{t}=xt$$ that $$0\le t\le x$$ means that $$0\le \widetilde{t}\le x^2$$ using $$\widetilde{t}/x=t$$. Plugging this all in, we get $$\int_0^x f(x,t)dt=\int_0^{x^2} \frac{f(x,\widetilde{t}/x)}{x}d\widetilde{t}.$$
• What about the argument of $f$? Feb 8 '19 at 1:31