Treating a definite integral as an indefinite integral during $u$-substitution

Question

When using substitution on a definite integral, I understand that its limits of integration need to be changed to fit the substitution.

But can I not do the computation as if it's an indefinite integral, then in the end use the original limits of integration?

Answer 1

Yes you can, but you have to return substitution at the end. For example, if your substitution is x=t+1 when you calculate indefinite integral I(t), yo have to convert it to I(x) (substitute every t with t=x-1) and then use limits from the begining.

Answer 2

Not sure I understand what you mean. Take $$ \int_1^2 x dx =3/2\ . $$ Now set $x=y^2$. So $dx=2y dy$. The indefinite integral in $y$ reads $$ \int dy (2y^3)=y^4/2\ . $$ If you compute it between the original limits $1,2$ [without reverting back to $x$] you get $2^4/2-1^4/2=15/2$, which clearly is incorrect.

Answer 3

Yes, if you solve for the indefinite integral and substitute whatever variable you used for the integration and then use the original bounds, then evaluating the integral will be the same.

Sometimes, it is easier to do what you just said, especially when you have to use multiple substitution techniques (this is often the case in trigonometric integrals).