Showing $8.9<\int_3^5 \sqrt{4+x^2} \, \mathrm d x < 9$

Question

I am asked to show that

$$8.9<\int_3^5 \sqrt{4+x^2} \, \mathrm d x < 9$$

I tried computing the integral but I end up with

$$\frac 5 2 \sqrt{29} - \frac 3 2 \sqrt{13} + 2 \log{\left(\frac {5 + \sqrt{29}} {3+\sqrt{13}}\right)}$$

which isn't really easy to approximate. I was thinking there might be a clever way of bounding the integrand with two functions that are easy enough to integrate. I was thinking Taylor series but I'm not sure how I can ensure that gives me an actual bound as opposed to a mere approximation.

Answer 1

Let $f(x)=\sqrt{4+x^2}$. Then $f$ is convex in $[3,5]$ and so the integral $I$ is less than area of the trapezoid above the graph: $$ I < \frac{f(3)+f(5)}{2}\cdot (5-3) = f(3)+f(5) = \sqrt{13}+\sqrt{29} \approx 8.99 < 9 $$

Answer 2

For $\,3<x<5\,$ we have $\,3.55+0.9(x-3)<\sqrt{4+x^2}<3.61+0.89(x-3)\,$ .

It's easily proofed by setting $\,x:=z+4\,$ with $\,-1<z<1\,$ and by the inequality conversion to

$-0.0321-0.01(1+z)-0.2079(1-z^2)<0<0.1875+0.01(1-z)+0.19z^2\,$.

Integration for $\,x\,$ from $\,3\,$ to $\,5\,$ proofs the assumption.

Answer 3

$$\int_{3}^{5}\sqrt{4+x^2}\,dx = \int_{0}^{1}\underbrace{\sqrt{4+(4-x)^2}+\sqrt{4+(4+x)^2}}_{f(x)}\,dx $$ where $f(x)$ is increasing and convex on the interval $[0,1]$, going from $f(0)=4\sqrt{5}$ to $f(1)=\sqrt{13}+\sqrt{29}$. By the Hermite-Hadamard inequality it follows that the wanted integral is between $f\left(\frac{1}{2}\right)=\frac{\sqrt{65}+\sqrt{97}}{2}\geq 8.9555\ldots$ and $\frac{f(0)+f(1)}{2}=\frac{4\sqrt{5}+\sqrt{13}+\sqrt{29}}{2}\leq 8.9675$.

Since $f(x)$ essentially has a quadratic behaviour on $[0,1]$, the approximation given by Simpson's rule $\frac{f(0)+4f\left(\frac{1}{2}\right)+f(1)}{6}\approx \color{green}{8.9595}4$ is pretty accurate.