Question

Consider the equation \[ \frac{dy}{dx} = \frac{1}{x} \] The value of the integral \(\int_{1}^{2} dy\) using trapezoidal method and interval \(h = 0.25\) is ________. (rounded off to two decimal places)

Hint:

Use the composite trapezoidal rule on \(f(x)=1/x\) with 4 strips of width 0.25 between \(x=1\) and \(x=2\).

Answer 1

Correct Answer: 0.7

Step 1: Set up the integral to be evaluated numerically.
The equation \(\frac{dy}{dx} = \frac{1}{x}\) means \(y\) is the antiderivative of \(f(x) = \frac{1}{x}\), so the integral asked for is
\[ \int_{1}^{2} dy = \int_{1}^{2} \frac{1}{x}\, dx \]
We are told to find this using the trapezoidal rule with step size \(h = 0.25\), instead of solving it exactly.

Step 2: Build the table of function values.
With \(h = 0.25\) over \([1, 2]\), the nodes are \(x_0 = 1, x_1 = 1.25, x_2 = 1.5, x_3 = 1.75, x_4 = 2\).
Evaluate \(f(x) = 1/x\) at each node:
\[ f(1) = 1,\ f(1.25) = 0.8,\ f(1.5) = 0.66667,\ f(1.75) = 0.57143,\ f(2) = 0.5 \]

Step 3: Apply the trapezoidal rule.
The composite trapezoidal rule for \(n\) intervals is
\[ \int_{a}^{b} f(x)\,dx \approx \frac{h}{2}\left[f(x_0) + f(x_n) + 2\sum_{i=1}^{n-1} f(x_i)\right] \]
The two end values are \(f(x_0) = 1\) and \(f(x_4) = 0.5\), and the interior values are \(f(x_1), f(x_2), f(x_3)\).
\[ \sum_{i=1}^{3} f(x_i) = 0.8 + 0.66667 + 0.57143 = 2.03810 \]

Step 4: Substitute and compute.
\[ \int_{1}^{2} \frac{1}{x}\,dx \approx \frac{0.25}{2}\left[1 + 0.5 + 2(2.03810)\right] \]
\[ = 0.125 \times [1.5 + 4.07620] = 0.125 \times 5.57620 = 0.69703 \]
As a check, the exact value is \(\ln 2 \approx 0.6931\); the trapezoidal estimate is a little higher because \(1/x\) curves downward (is convex) on this interval, which is the usual behavior of this method for a convex function.

Final Answer:
Rounded off to two decimal places,
\[ \boxed{0.70} \]