Dtyjlui

I need to calculate:
$$
lim_{xto infty} int_x^{x+1} frac{t^2+1}{t^2+20t+8}, dt
$$ 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.

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

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

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.

This is an easy consequence of the definition of limit.

Note that the integrand $f(t) $ tends to $1$ as $ttoinfty $ 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) tomathbb {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 $xto infty $. Then $int_{x} ^{x+1}f(t),dtto L$ as $xto infty $.

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