Question

The table below gives the values of \( f(x) \) at five equidistant points of \( x \):

x	0	0.5	1.0	1.5	2.0
f(x)	0	0.25	1.0	2.25	4.0

Then the approximate value of \( \int_0^2 f(x) \, dx \) by Trapezoidal Rule is:

• A. \(1.75\)
• B. \(2\)
• C. \(2.5\)
• D. \(2.75\)

Hint:

The Trapezoidal Rule is especially useful when values of \( f(x) \) are given at equidistant points. Always remember the endpoints are taken once, and interior points are doubled in the formula.

Answer 1

The Correct Option is D

Using the Trapezoidal Rule formula: \[ \int_a^b f(x)\, dx \approx \frac{h}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4) \right] \] Here, \( h = 0.5 \) (since \( x \) increments by 0.5), \( x_0 = 0, x_1 = 0.5, x_2 = 1, x_3 = 1.5, x_4 = 2 \), \( f(x_0) = 0,\ f(x_1) = 0.25,\ f(x_2) = 1,\ f(x_3) = 2.25,\ f(x_4) = 4 \) Substituting into the formula: \[ \int_0^2 f(x)\, dx \approx \frac{0.5}{2} \left[ 0 + 2(0.25) + 2(1) + 2(2.25) + 4 \right] \] \[ = \frac{0.5}{2} \left[ 0 + 0.5 + 2 + 4.5 + 4 \right] = \frac{0.5}{2} \cdot 11 = 0.25 \cdot 11 = {2.75} \]