How to prove $\frac{1}{2}\leq\int_0^2 \frac{dx}{2+x^2}\leq\frac{5}{6}$?

Question

How to prove the following? $$\frac{1}{2}\leq\int_0^2 \frac{dx}{2+x^2}\leq\frac{5}{6}$$

Ok I know we can directly integrate the function.The answer is given by Wolfram Alpha

But is there any shorter/elegant method to obtain the left and right bounds directly by squeeze theorem (i.e. by integrating two function in between which the given function lies) ?

@ParclyTaxel Oh! It's from my maths study material.It's a question in itself. — , Aug 13 '16 at 12:02
http://math.stackexchange.com/questions/1405602/prove-that-1-frac12-leq-int-02-fracdx2x2-leq-frac56br/1405610#1405610 — juantheron, Aug 13 '16 at 12:09

Michael Hardy · Accepted Answer · 2016-08-13T12:18:10.980

2

If $0\le x\le 1$ then $\dfrac 1 3 \le \dfrac 1 {2+1^2} \le \dfrac 1 {2+x^2} \le \dfrac 1 {2+0^2} = \dfrac 1 2$.

If $1\le x \le 2$ then $\dfrac 1 6 \le\dfrac 1 {2+x^2} \le \dfrac 1 3 $.

Therefore $$ \frac 1 3 + \frac 1 6 \le \int_0^1 \frac{dx}{2+x^2} + \int_1^2 \frac{dx}{2+x^2} \le \frac 1 2 + \frac 1 3. $$

As for Wolfram Alpha, notice that the integral is easily evaluated in closed form: \begin{align} & \int_0^2 \frac{dx}{2+x^2} = \frac 1 2 \int_0^2 \frac{dx}{1 + \left(\frac x {\sqrt 2}\right)^2} = \frac{\sqrt 2} 2 \int_0^2 \frac{dx/\sqrt 2}{1+\left( \frac x {\sqrt2} \right)^2} = \frac {\sqrt 2} 2 \int_0^{\sqrt 2} \frac{du}{1+u^2} \\[10pt] = {} & \frac{\sqrt 2} 2 \left( \arctan \sqrt 2 - \arctan 0 \right) = \frac{\sqrt 2} 2 \arctan \sqrt 2. \end{align}

edited Aug 13 '16 at 12:18

answered Aug 13 '16 at 12:05

Michael Hardy

1

That was really clever.BTW how did you identify it?Or is it a mainstream problem?Have you done similar problems before? – Aug 13 '16 at 12:08
It is fairly standard to say that if $$ m \le f(x) \le M $$ then $$ m(b-a) \le \int_a^b f(x),dx \le M(b-a). $$ – Michael Hardy Aug 13 '16 at 12:10
2

And if the bounds for the full interval are to coarse, split the interval. Thus essentially arriving at Riemann/Darboux upper and lower sums. – Lutz Lehmann Aug 13 '16 at 12:14

score 2 · Answer 2 · answered Aug 13 '16 at 12:06

2

On $[0,1]$ you get $$ \frac13\le\frac1{2+x^2}\le\frac12 $$ and on $[1,2]$ $$ \frac16\le\frac1{2+x^2}\le\frac13 $$ The combination of the estimates for both intervals gives exactly the requested bound.

answered Aug 13 '16 at 12:06

Lutz Lehmann

126,666

score 0 · Answer 3 · answered Aug 13 '16 at 14:48

If we don't want use the closed form of the integral, we can see that $$\int_{0}^{2}\frac{dx}{2+x^{2}}\stackrel{x=2u}{=}\int_{0}^{1}\frac{du}{1+2u^{2}} $$ and $$\int_{0}^{2}\frac{dx}{2+x^{2}}=\int_{0}^{1}\frac{du}{1+2u^{2}}\geq\int_{0}^{1}\frac{du}{1+2u}=\frac{1}{2}\log\left(3\right)>\frac{1}{2},$$ $$\int_{0}^{1}\frac{du}{1+2u^{2}}\leq\int_{0}^{1}\frac{du}{1+u^{2}}=\frac{\pi}{4}<\frac{5}{6}.$$

How to prove $\frac{1}{2}\leq\int_0^2 \frac{dx}{2+x^2}\leq\frac{5}{6}$?

3 Answers3