find an upper/lower bound for the integral

• Aug 27th 2006, 12:26 PM
FLTR
find an upper/lower bound for the integral
Hello, am a complete novice attempting an on-line course, can you show me how to work this problem? Thanks for your time.

Find an upper and lower bound for the integral $\int_{0}^{1}\frac{1}{x+2}dx$ using the comparison properties of integrals.
• Aug 27th 2006, 04:59 PM
ThePerfectHacker
Quote:

Originally Posted by FLTR
Hello, am a complete novice attempting an on-line course, can you show me how to work this problem? Thanks for your time.

Find an upper and lower bound for the integral $\int_{0}^{1}\frac{1}{x+2}dx$ using the comparison properties of integrals.

Maybe this will help, for $x>0$
$0<\frac{1}{x+2}<\frac{1}{x}$
Then,
$0<\int_0^1 \frac{dx}{x+2} < \int_{0^+}^1 \frac{dx}{x}=\ln x$
• Aug 27th 2006, 08:23 PM
Soroban
Hello, FLTR!

Quote:

Find an upper and lower bound for the integral $\int_{0}^{1}\frac{1}{x+2}dx$
using the comparison properties of integrals.

Since $0 \leq x \leq 1:\;\;2\:\leq \;x+2\:\leq \:3\quad\Rightarrow\quad \frac{1}{3} \,\leq \,\frac{1}{x+2} \,\leq \,\frac{1}{2}$

Hence: . $\int^1_0\!\frac{1}{3}\,dx \;\leq \;\int^1_0\!\!\frac{dx}{x+2} \; \leq \; \int^1_0\!\frac{1}{2}\,dx$

Therefore: . $\frac{1}{3} \;\leq \;\int^1_0\!\!\frac{dx}{x+2} \;\leq \;\frac{1}{2}$

• Aug 28th 2006, 02:50 PM
FLTR
Thanks for the examples. I should be able to work the remaining exercises, at least on this type of problem!