Question

re-evaluating the integrals limits

Answer 1

\[\int\limits_{0}^{10}\frac{ dx }{ 1+x }\]

Answer 2

I can't remember how to re-evaluate the limits after u-substitution

Answer 3

$$\int_0^{10}\frac{dx}{1+x}=\left[\log(1+x)\right]_0^{10}=\log11-\log1=\log11$$

Answer 4

You simply plug in the values as x. If u = x+1 then for the bottom limit you have 0+1 = 1 and for the top you have 10+1 = 11 giving:
\[\int\limits_1^{11} \frac{du}{u}\]

Answer 5

so, u=1+x then plug in the x valvues to get the new limits?