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

Here $y$ satisfies $dy/dx = 1/x$, so the quantity asked for is just the definite integral $\int_1^2 (1/x)\,dx$, evaluated approximately by adding up the areas of four small trapezoids of width $h = 0.25$ instead of integrating exactly.

Split $[1, 2]$ into four equal strips of width $0.25$: $[1, 1.25]$, $[1.25, 1.5]$, $[1.5, 1.75]$, $[1.75, 2]$. First list $f(x) = 1/x$ at the five boundary points:

• $f(1) = 1.00000$
• $f(1.25) = 0.80000$
• $f(1.5) = 0.66667$
• $f(1.75) = 0.57143$
• $f(2) = 0.50000$

Each strip is approximated by a trapezoid, and its area is the average of the two end heights times the width $h$:

1. Strip 1: $\frac{0.25}{2}(1.00000 + 0.80000) = 0.22500$
2. Strip 2: $\frac{0.25}{2}(0.80000 + 0.66667) = 0.18333$
3. Strip 3: $\frac{0.25}{2}(0.66667 + 0.57143) = 0.15476$
4. Strip 4: $\frac{0.25}{2}(0.57143 + 0.50000) = 0.13393$

Add up the four strip areas to get the total integral estimate:

$$ 0.22500 + 0.18333 + 0.15476 + 0.13393 = 0.69702 $$

Let's summarize:

• The trapezoidal rule replaces the curve $1/x$ on each small strip with a straight line joining its two end points, then adds up the resulting trapezoid areas.
• The exact value of $\int_1^2 (1/x)\,dx$ is $\ln 2 \approx 0.6931$, so this four-strip estimate is already close, off by less than $0.004$.

Rounded off to two decimal places, the value of the integral is $0.70$.