Question

Using Trapezoidal rule with one interval, find the approximate value of \(\displaystyle \int_{1}^{2}\frac{dx}{1+x^{2}}\) (rounded off to 2 decimal places).

Hint:

For a single subinterval, the trapezoidal rule is simply the average of the endpoint values times the width: \(\dfrac{b-a}{2}[f(a)+f(b)]\).

Answer 1

Step 1: Identify nodes and step size.
With one interval on \([1,2]\): \(a=1,\; b=2,\; h=b-a=1\).
Let \(f(x)=\dfrac{1}{1+x^{2}}\). 

Step 2: Trapezoidal rule (single interval).
\[ \int_{a}^{b} f(x)\,dx \approx \frac{h}{2}\,\big[f(a)+f(b)\big]. \] 

Step 3: Evaluate \(f\) at the endpoints and compute.
\(f(1)=\dfrac{1}{1+1}=\dfrac{1}{2}=0.5,\qquad f(2)=\dfrac{1}{1+4}=\dfrac{1}{5}=0.2.\)
\[ \Rightarrow\ \int_{1}^{2}\frac{dx}{1+x^{2}} \approx \frac{1}{2}\,(0.5+0.2)=\frac{1}{2}\times 0.7=0.35. \] Rounded to two decimals: \(\boxed{0.35}\).