Calculating limit of definite integral

Question

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.

Answer 1

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.

Answer 2

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?

Answer 3

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

Answer 4

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.

Answer 5

This is an easy consequence of the definition of limit.

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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 $$1-\epsilon <f(t) <1+\epsilon $$ 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 $.

Answer 6

Use the estimate: $$\frac{t-10}{t+10}<\frac{t^2+1}{t^2+20t+8}<\frac t{t+10}, \ t>1 \Rightarrow \\ \int_x^{x+1} \frac{t-10}{t+10}dt<I(x)<\int_x^{x+1} \frac t{t+10}dt \Rightarrow \\ 1-20\ln \frac{x+11}{x+10}<I(x)<1-10\ln \frac{x+11}{x+10} \Rightarrow \\ \lim_{x\to\infty} \left(1-20\ln \frac{x+11}{x+10}\right)\le \lim_{x\to\infty} I(x) \le \lim_{x\to\infty} \left(1-10\ln \frac{x+11}{x+10}\right) \Rightarrow \\ \lim_{x\to\infty} I(x)=\lim_{x\to\infty} \int_x^{x+1}\frac{t^2+1}{t^2+20t+8}dt=1.$$