# Why is it that, when using the u-substitution rule, you apply the the function u to the upper and lower limits?

I was taught that we apply the function u to the upper and lower bounds of the definite integral, then use those as the new upper and lower bounds to evaluate the integral. Indeed, this works and is the only way I can get the right answer on question involving substitution. But why do we have to do this?

Why can't you just find the indefinite integral, then use the 2nd part of the fundamental theorem of calculus (i.e. evaluate the antiderivative at the lower bound and minus that from the value of the antiderivative at the upper bound) to solve the problem from there? Why do you have to change the upper and lower limits as you make the u-substitution?

Firstly, I'm lacking the intuition as to why you actually have to do that. Secondly, I'm confused as to why doing what I described doesn't work, as it seems like it should.

Any help is greatly appreciated.

• You can do the substitution in an indefinite integral, and then use the answer you get there to find the definite one. That requires: (i) doing the indefinite integral separately, because it is not the same as the definite integral; and (ii) remembering to go back to the original variable when you are done integrating, so that you have a function of the same variable as you started with. In contrast, if you do the substitution into a definite integral, you have to change the limits to maintain equality, but then you don’t need to go back to the original variable. Your choice. – Arturo Magidin Dec 3 '18 at 21:20
• Why do you change the limits? Because you are working on a different function. Think of $\sin(2x)$, whose graph is the graph of sine, but “twice as fast” (completes one full period in $\pi$). If you are integrating this from $0$ to $\frac{\pi}{2}$, you are only integrating half the period. If you switch to $\sin(u)$ with $u=2x$, you need to account for this fact (if you integrate $\sin(u)$ from $0$ to $\pi/2$, you are only integrating over a quarter of the period). – Arturo Magidin Dec 3 '18 at 21:22

## 1 Answer

Consider the integral $$I=\int_{x_1}^{x_2}f(g(x))g'(x)dx$$ It may be helpful if we note that we are allowed to write the integral as $$I=\int_{x=x_1}^{x=x_2}f(g(x))g'(x)dx$$ First of all, we can apply the substitution $$u=g(x)$$, which gives $$f(g(x))g'(x)dx=f(u)du$$. But the integral still has bounds in terms of $$x$$. To solve this problem, we note that $$g^{-1}(u)=x$$ Hence we can say that, if $$x=x_1$$, then $$g^{-1}(u_1)=x_1\\u_1=g(x_1)$$ The same applies for the upper bound: $$u_2=g(x_2)$$ Finally, we can write our integral as $$I=\int_{u=u_1}^{u=u_2}f(u)du$$ Or simply $$I=\int_{g(x_1)}^{g(x_2)}f(u)du$$

To verify this we assume that $$F'(x)=f(x)$$.

Hence we start on the indefinite integral $$J(x)=\int f(g(x))g'(x)dx$$ Again, we preform $$u=g(x)$$: $$J(x)=\int f(u)du$$ Which we know: (I am omitting the constant of integration FYI) $$J(x)=F(u)$$ But this is still in terms of $$u$$. Fortunately, this is an easy fix: $$J(x)=F(g(x))$$ And by the fundamental theorem of calculus, we have that $$I=J(x_2)-J(x_1)$$ Which of course is \begin{align} I=&F(g(x_2))-F(g(x_1))\\ =&F(u_2)-F(u_1)\\ \end{align} Hence we have our conclusion: $$\int_{x_1}^{x_2}f(g(x))g'(x)dx=\int_{g(x_1)}^{g(x_2)}f(u)du$$