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When you use "u-substitution" with a definite integral you have to find the new limits for the integral.

If the lower limit in the integral is $a$ and the higher limit $b$ (and o is for original and p for after u substitution) and

$u = f(x)$

then

$pa = f(oa)$ and $pb = (ob)$

Sometimes this results in $pb<pa$

e.g. $u = 10 - x $, $oa = -10$ and $ob = 6$ then $pa = 20$ and $pb = 4$

What happens in a situation like this? Do you swap the limits and make the limit with the higher value the higher limit?

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    $\begingroup$ No. If you swap the limits the sign of the integral change. $\endgroup$ – Emilio Novati Feb 17 '17 at 11:06
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You can use the following property of integrals in this situation: $$\int \limits_{a}^{b} f(x) \,dx = - \int \limits_{b}^{a} f(x) \,dx.$$ That is, if you switch the upper and lower limit, you have to introduce a factor of $-1$ into the integrand.

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