I dont know if substitution would work here, but not 100% sure. I know you could easily open up the brackets to find the integral to be:
I want to use u-substitution in my self-study-case.
If I have equation: and then I substitute:
And then I plugin the u:
And finally plugin the original x:
Whats wrong with my substitution. I want to use Is this substitution even possible?
If f is a function of u and u is a function of x, then df/dx= (df/du)(du/dx)
When differentiating, we are free to just go ahead and multiply that "du/dx" in.
If we have an integral , that is the same as [tex]\inf f(u)du[/itex]. Unfortunately, if that "du/dx" is not inside the integral to begin with, we cannot just put it in: we cannot, for example, say "if so that so because you cannot cancel an x outside the integral with an x inside the integral.
You CAN cancel constants inside and outside the integral because [tex]\int c f(x)dx= c\int f(x)dx[/itex] for c a constant. If that problem were , then we could say "let u= 3x- 1 so du= 3dx. Then .
But in general we cannot make the subsitution u= f(x) unless du/dx is alread inside the integral. That's one thing that make integration so much harder than differentiation.