# Thread: Use upper and lower sum to show that

1. ## Use upper and lower sum to show that

0.6 < $\displaystyle \int _0\,^1\!dx/(1+x^2) <1$

I am not sure what the partition is, to find the upper and lower sum, we need the partition and sub interval to do it. Can someone clear it ?

2. $\displaystyle 0.6< \int_0^1\frac{dx}{1+x^2} <1 \implies 0.6<\tan^{-1}x+c <1$

3. Originally Posted by pickslides
$\displaystyle 0.6< \int_0^1\frac{dx}{1+x^2} <1 \implies 0.6<\tan^{-1}x+c <1$

What! where is the tan from? Don't we need to find the lower sum and upper sum?

4. Originally Posted by 450081592
What! where is the tan from?
$\displaystyle \int_0^1\frac{dx}{1+x^2} =\left[\tan^{-1}x \right]_0^1$

Just thought it would come in helpful.

5. Originally Posted by pickslides
$\displaystyle \int_0^1\frac{dx}{1+x^2} =\left[\tan^{-1}x \right]_0^1$

Just thought it would come in helpful.

I still don't understand where tan come from? which one is the lower sum and upper sum? We havent covered how to find the integration of a function, Is the first time I deal with interation, so can you explain more thanks