Upper and lower bounds of int [0,1] 1/(x+2) dx

[confused_07]

Find the upper and lower bounds for the integral int [0,1] 1/(x+2) dx using the comparison properties of integrals.

So, my text book states the comparison properties, and then gives only ONE example on how to do it. In this example though, they use three functions. How do you do this with only one?

 

[pka]

\(\displaystyle \L \frac{1}{3} < \frac{1}{{x + 2}} < \frac{1}{{x + 1}}\quad \Rightarrow \quad \int\limits_{0}^1 {\frac{1}{3}} < \int\limits_0^1 {\frac{1}{{x + 2}}} < \int\limits_0^1 {\frac{1}{{x + 1}}}\)

 

[confused_07]

So you just plug in the [0,1] into the equation to get your bounds?

 

[pka]

So you just plug in the [0,1] into the equation to get your bounds?

No.

\(\displaystyle \L \int\limits_0^1 {\frac{1}{3}} = \frac{1}{3}\quad \& \quad \int\limits_0^1 {\frac{1}{{x + 1}}} = \ln (2).\)

 

[confused_07]

Sorry should have been more clear. I was asking how you got the 1/3 and the 1/(x+1). I see by plugging in the '1', you got 1/3....

 

[pka]

\(\displaystyle \L x \in [0,1] \Rightarrow \quad x + 1 \le x + 2 \le 3 \Rightarrow \quad \frac{1}{3} \le \frac{1}{{x + 2}} \le \frac{1}{{x + 1}}\)