Bounds for an integral

Question

I am trying to show that $$\frac{1}{5} < \int_5^8 \frac{2x-7}{2x+5} dx <1$$ since for the integral $$5\le x \le 8 \rightarrow 15\le 2x+5 \le 21$$$$-\frac{12}{15}\le-\frac{12}{2x+5}\le -\frac{12}{21}\rightarrow \frac{3}{15}\le 1- \frac{12}{2x+5}\le \frac{3}{7}$$ By taking integral $$\frac{3}{5}\le \int_5^8 \left(1-\frac{12}{2x+5}\right)dx\le\frac{9}{7}$$ Whilee the first one could be reduced to what is needed, how can I deal with the right bound?

Answer 1

Tangent line to $y=\frac{2x-7}{2x+5}$ at $x=5$ and $x=7$ are: $y=\frac{8}{75}x-\frac13$ and $y=\frac{24}{361}x-\frac{35}{361}$, respectively. Refer to the graph:

$\hspace{0.5cm}$

Hence: $$\int_5^8\frac{2x-7}{2x+5}dx<\int_5^6 \left(\frac{8}{75}x-\frac13\right)dx+\int_6^8 \left(\frac{24}{361}x-\frac{35}{361}\right)dx\approx \\ 0.2533333333+0.736842105=0.9901754383<1.$$

Answer 2

We wish to show that $$\int_5^8{\frac{2x-7}{2x+5}dx}<1.$$ Let $g:(0,\infty)\to\mathbf R$ be given by $$g(x)=1-\int_5^x{1-\frac{12}{2w+5}dw}.$$ Then $$g'(x)={12\over{2x+5}}-1$$ and $$g''(x)={-24\over{(2x+5)^2}}<0\,\, \forall x.$$ Thus, $g$ is concave. Also, $g'=0$ at $x=3.5$, so that $g$ has its turning point $\left( g'(3)>0\,\text{and}\,g'(4)<0 \right)$ at $x=3.5$, where $g=2.5-6\log \frac 54>0$ since $4e^{5/12}>5.$ Thus, $g$ has just two roots.

Clearly, one of the roots must be $<3.5$, so that we need only consider the other root $>3.5$ (since we're interested only in how this function behaves in $[5,8]$). I claim that the other root is also $>8.$ Now we have that $$g(x)=6-x-6\log(2x+5),$$ so that $$g(8)=6\log \frac75 -2>0.$$ That last inequality holds because clearly $7>5e^{1/3}.$ Also, $$g(20)=-14+6\log 3<-14+12=-2<0,$$ since $3<e^2=(2+\varepsilon)^2=4+\delta$, where $1/2<\varepsilon<1$ and $\delta=4\varepsilon+\varepsilon^2.$

Consequently, the other root of $g$ must be in $(8,20)$, so that $g>0$ in $[5,8]$ (since $g(5)=1>0$). The result follows by setting $x=8$ in $g(x)>0.\blacksquare$

Answer 3

$f(x)=\frac{2x-7}{2x+5}$ is a concave function on $[5,8]$, hence by the Hermite-Hadamard inequality the wanted integral is bounded between $$ \frac{1}{2}f(5)+f(6)+f(7)+\frac{1}{2}f(8) = \frac{11043}{11305} > \frac{42}{43}$$ and $$ f\left(\tfrac{11}{2}\right)+f\left(\tfrac{13}{2}\right)+f\left(\tfrac{15}{2}\right) = \frac{59}{60}<1.$$ As an alternative, one may notice that the exact value is given by $3-6\log\left(1+\frac{2}{5}\right)$, and by Padé approximants $\log(x+1)$ is bounded between $\frac{x}{1+\frac{x}{2}}$ and $\frac{x+\frac{x^2}{6}}{1+\frac{2x}{3}}$ over $[0,1]$. This gives that the wanted integral is between $\frac{93}{95}$ and $1$.