# Calculating limit of definite integral

I need to calculate: $$\lim_{x\to \infty} \int_x^{x+1} \frac{t^2+1}{t^2+20t+8}\, dt$$ The result should be $$1$$.

Is there a quicker way than calculating the primitive function?

I thought about seperating to $$\int_0^{x+1} -\int_0^x$$ but still can't think of the solution.

• The searched limit is equal to $1$ Jan 31, 2019 at 15:13
• Your chosen title is inappropriate, since there is no improper integral in your question. Jan 31, 2019 at 15:27
• Can you find upper and lower bounds of the integrand? Jan 31, 2019 at 22:11

Why not apply the MVT? The integral is $$F(x+1)-F(x)=\frac{F(x+1)-F(x)}{x+1-x}=F'(x_0)$$ for an $$x_0$$ between $$x$$ and $$x+1$$. Now $$F'(x)$$ certainly converges to $$1$$ if $$x$$ tends to infinity.

Note that\begin{align}\int_x^{x+1}\frac{t^2+1}{t^2+20t+8}\,\mathrm dt&=\int_x^{x+1}1\,\mathrm dt+\int_x^{x+1}\frac{-20t+7}{t^2+20t+8}\,\mathrm dt\\&=1-20\int_x^{x+1}\frac{t-\frac7{20}}{t^2+20t+8}\,\mathrm dt.\end{align}So, all that remains to be proved is that$$\lim_{x\to\infty}\int_x^{x+1}\frac{t-\frac7{20}}{t^2+20t+8}\,\mathrm dt=0.$$Can you take it from here?

• Oh i thought about it too, but wasn't sure how to continue. so i guess i prove it with the squeeze theorem like the answer of mathcounterexamples?
– Ido
Jan 31, 2019 at 15:53
• That's what I would do, yes. Jan 31, 2019 at 16:08
• I've got one more question - I now understand the idea of using the squeeze theorem, but is it legitimate to argue that when $x$ goes to $\infty$ than the "t" function goes to $1$, and claim that therefore the result is the limit of $1*(x+1-x)=1$? thanks
– Ido
Jan 31, 2019 at 16:30
• A proof can be found along these lines, yes. Jan 31, 2019 at 17:42
• @Ido: see my answer for a "proof along these lines". Feb 1, 2019 at 17:42

For $$t > 8$$ you have:

$$0 \le 1 - \frac{t^2+1}{t^2+20t+8} = \frac{20t+7}{t^2+20t+8} \le \frac{20t+8}{t^2} \le \frac{21}{t}$$

Hence integrating those inequalities on $$[x,x+1]$$:

$$0 \le 1- \int_x^{x+1} \frac{t^2+1}{t^2+20t+8}dt \le 21 \int_x^{x+1} \frac{dt}{t} \le \frac{21}{x}$$

proving that $$\lim_{x\to \infty} \int_x^{x+1} \frac{t^2+1}{t^2+20t+8}dt =1$$.

• I see, thank you very much :)
– Ido
Jan 31, 2019 at 15:54

For large values of $$x$$, the graph of $$f(t) = \frac{t^2+1}{t^2+20t+8}$$ [you can show this by finding $$f'(t)$$ and show that $$f'(t)$$ approaches zero as $$t$$ approaches to infinity] is approximately horizontal, so the area represented by the integral is approximately a rectangle with width $$(x + 1) - x = 1$$ and height $$f(x) = \frac{x^2+1}{x^2+20x+8}$$, which approaches 1 as $$x$$ approaches infinity.

• That's a visualisation of the approach via MVT. Feb 1, 2019 at 12:09

This is an easy consequence of the definition of limit.

Note that the integrand $$f(t)$$ tends to $$1$$ as $$t\to\infty$$ and hence corresponding to every $$\epsilon >0$$ we have a corresponding $$M_{\epsilon} >0$$ such that $$1-\epsilon whenever $$t>M_\epsilon$$. Let $$x>M_\epsilon$$ and then integrating the above inequality with respect to $$t$$ in interval $$[x, x+1]$$ we get $$1-\epsilon <\int_{x} ^{x+1}f(t)\,dt< 1+\epsilon$$ whenever $$x>M_\epsilon$$ and thus by definition the desired limit is $$1$$.

There is nothing special about the integrand and its limit and what we have proved above can be summarized as the following

Lemma: Let $$f:[a, \infty) \to\mathbb {R}$$ be a function which is Riemann integrable on every interval of type $$[a, b]$$ with $$b>a$$ and let $$f(x) \to L$$ as $$x\to \infty$$. Then $$\int_{x} ^{x+1}f(t)\,dt\to L$$ as $$x\to \infty$$.

Use the estimate: $$\frac{t-10}{t+10}<\frac{t^2+1}{t^2+20t+8}<\frac t{t+10}, \ t>1 \Rightarrow \\ \int_x^{x+1} \frac{t-10}{t+10}dt