I have this following Integral:
$$\int_0^2 (2x-1)^3 dx$$
I want to integrate it using u-substitution, like that:
$$u = 2x-1 $$ $${\frac{du}{dx}(2x + 1)} = 2$$ $$ {du = (2){dx}}$$ $$1/2\int (u)^3{du} $$
So far so good. After that I've got to change the boundaries in the integral, there comes the problem: I can't visualize this change graphically. I know that when
$$x = 0$$ $$u = -1$$
but I mean, if I plot $$ (u)^3 $$ It doesn't have the same graph as $$(2x-1)^3$$
How can they be equal?
I am sorry If it is a naive question and if I couldn't express myself right, I can clarify in the comments though
Thanks in advance
Simplistic approach to address the question in your title:Let's take a look at a MUCH easier situation. Consider $y=4-x^2$ and consider the area under the curve from $x=-2$ to $x=2$. I think you can figure that out, right? Now let's translate this area $2$ units to the right. The function then becomes $y=4-(x-2)^2$. If I want to calculate the SAME area under the curve, how can I not have to move along the x-values of integration? Since the area literally has moved $2$ units to the right, in order to deal with the same region, there is no other way that I move those $x$ values $2$ units to the write as well. Now if you would perform a substitution on the $(x-2)^2$, by setting $x-2=t$, the idea of the substitution rule with respect to "tagging the boundary values along" becomes evident