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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?

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4 Answers 4

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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.

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Let's say you are computing $\displaystyle\int_2^7\frac{x^2}{x^6+1}\,\mathrm dx$ by first treating it as an indefinite integral using the substitution $\displaystyle u=x^3.$

  1. This working is valid, since compared to the usual procedure it merely delays changing the integration limits: $$\int\frac{x^2}{x^6+1}\,\mathrm dx=\ldots=\frac13\arctan(u)+C.\\ \int_2^7\frac{x^2}{x^6+1}\,\mathrm dx=\left[\frac13\arctan(u)-\frac13\arctan(u)\right]_{u=2^3}^{u=7^3}.$$

  2. This working is also valid, because it is clearly equivalent to the above, or alternatively by invoking the Fundamental Theorem of Calculus: $$\int\frac{x^2}{x^6+1}\,\mathrm dx=\ldots=\frac13\arctan(u)+C=\frac13\arctan(x^3)+C.\\ \int_2^7\frac{x^2}{x^6+1}\,\mathrm dx=\left[\frac13\arctan(x^3)-\frac13\arctan(x^3)\right]_{x=2}^{x=7}.$$

    On a related note: a shorthand notation for integration-by-substitution that I commonly see on this website is $$\int_a^b f\big(g(x)\big)\, g'(x)\,\mathrm dx :=\int_{\color\red{x=}a}^{\color\red{x=}b} f\big(g(x)\big)\,\mathrm d\big(g(x)\big).$$ Here, there is similarly no need to change the integration limits, since this presentation never actually changes the variable! Note that the "$\color{red}{x=}$" bit is essential, because on the right side the integration limit "$b$" alone must be read as "$g(x)=b$" instead of "$x=b$".

Caveat: when something like $\sqrt{\cos^2u}$ appears, the $u$-world integration limits are needed to determine whether to simplify it as $\cos u$ or $-\cos u.$ This is one of several reasons why I don't recommend your above suggestion.

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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.

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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).

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