improving estimation of integral Using this set of points of the form $(x,f(x))$: 
$$(1.0,1.54),(1.2,1.81),(1.4,2.15),(1.6,2.58),(1.8,3.11)$$
I estimated the integral using the extended trapezoid rule: 
$\int\limits_1^{1.8}f(x)dx\approx
1.79$ for $h=0.4$ and $\int\limits_1^{1.8}f(x)dx\approx
1.773$ for $h=0.2$.
Is it possible to get a more precise estimation of the integral using the above results? If so how.
 A: No. Briefly, while an error estimate can be computed, there is not enough information to determine if the technique is applicable or if the estimate is reliable. Moreover, the error estimate is comparable in magnitude to the error which the cautious user must assign to the trapezoidal sum.
Let $T$ denote the target value, i.e. the integral $T = \int_1^{1.8} f(x)dx$ and let $A_h = A_h(f)$ denote the trapezoidal sum for $f$ computed with uniform step size $h$. We can compute $A_h = 1.773$ and $A_{2h} = 1.79$ for $h = 0.1$. Richardson's error estimate is $$E_h = \frac{A_h - A_{2h}}{2^{p}-1} = -0.005\overline{6}, \quad (\text{for $ p = 2$})$$
where $p>0$ depends on the function $f$. There is not enough information to justify the application of Richardson's technique, identify the value of $p$ or determine if $E_h$ is an accurate approximation of the true error $T - A_h$. 
For this we need more points, so that several fractions of the form $$F_h = \frac{A_{2h} - A_{4h}}{A_h - A_{2h}}$$ can be computed. If these fractions converge monotonically to a positive value as $h \rightarrow 0_+$, then this is evidence of an asymptotic error expansion of the form $$T -A_h = \alpha h^p + \beta h^q + O(h^r), \quad 0 < p < q < r.$$
Moreover, this behavior would also indicate that rounding errors are much less significant than the discretization error, that the higher order terms are insignificant compared with dominant error term $\alpha h^p$ and that the error estimate $E_h$ is reliable. The order $p$ of the dominant error term would follow from the observed limit of $F_h$ since $F_h \rightarrow 2^p$ as $h \rightarrow 0_+$. The order $q$ of the secondary error term would follow from the observed convergence rate. If $p = 2$, then this would indicate that $f$ was several times differentiable on the entire interval. 
However, we do not have more points.
Returning to the information that we do have. In general, there is no reason to believe that the given values for $f$ are exact! At best, they are accurate to the number of significant figures shown. In this case we have three significant figures. In other words, the true value of, say, $f(1)$ is $f(1) = 1.54 \pm 0.005$, as any value in the interval $(1.535,1.545)$ rounds to midpoint $1.54$. The uncertainty on $f$ translates into an uncertainty on the computed value of the trapezoidal rule of $$A_h = 1.779 \pm 0.004.$$ 
The width of this interval is comparable to Richardson's error estimate $E_h$ in the smooth case where $p=2$. So, even in the best case, where $f$ is a smooth function, the step size $h = 0.2$ is so small as to quell the effect of the higher order terms, but so large that subtractive cancellation is not an issue and the error estimates would be reliable, the uncertainty on $f$ is enough to ruin our hopes.
We also need to know how accurately $f$ has been computed.
A: Under the assumption your points come from a polynomial, you can interpolate the polynomial to get $y=-0.260417 x^4 + 1.77083 x^3 - 3.23958 x^2 + 3.42917 x - 0.16$ and integrate that to get $1.73738$.  You can't do better.  It could be that the function has a huge bump around $x=1.1$ and the integral is much higher.
A: Any estimate of the integral is arbitrary because you just know the "function" by five points and interpolation is not unique.
You need to know some properties of the integrand to move on.
A: Assuming Simpson's rule is applicable, it can be utilized as a more accurate approximation relative to the trapezoidal rule, and is given by the following condensed formula:
$$S(P_n,f)=\frac h3\bigg[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)\bigg]$$
For $P_n$ being a uniform partition of cardinality $5$ (4 sub-intervals), and $h=\frac {b-a}{n}=\frac {1.8-1}{4}=0.2$, we have:
$$S(P_n,f)=\frac {0.2}{3}\bigg[1.54+4(1.81)+2(2.15)+4(2.58)+3.11\bigg]\approx1.76733$$
A: I think everybody is overthinking this. I agree with @YvesDaoust. The best you can do is to calculate the area under the curve that is given, and not presume to make any other assumptions as to what it might be or have been. I integrated this is in the complex plane and got the same answer you got with the trapezoidal method. To wit,
$$
z=x+if(x)\\
A=\frac{1}{2}\int_1^{1.8}\Im\{z^*\dot z \}~dx=1.773\  \text{(exactly)}$$
