GATE CS 2013 Q.23 (CS Q.23): Calculus

Function $f$ is known at the following points:

\begin{array}{c|ccccccccccc} x & 0 & 0.3 & 0.6 & 0.9 & 1.2 & 1.5 & 1.8 & 2.1 & 2.4 & 2.7 & 3.0 \\ \hline f(x) & 0 & 0.09 & 0.36 & 0.81 & 1.44 & 2.25 & 3.24 & 4.41 & 5.76 & 7.29 & 9.00 \end{array}

The value of $\int_0^3 f(x) dx$ computed using the trapezoidal rule is

8.983

9.003

9.017

9.045

Official answer and solution

Solution

Apply the composite Trapezoidal rule formula with uniform step size h = 0.3.

The composite Trapezoidal rule for numerical integration with step size $h = 0.3$ is:

I = \int_{x_0}^{x_n} f(x) dx \approx \frac{h}{2} \left[ (y_0 + y_n) + 2 \sum_{i=1}^{n-1} y_i \right]

Here $x_0 = 0, x_{10} = 3.0$ , and $n = 10$ .

Intermediate function values:

Sum of intermediate values:

\sum_{i=1}^9 y_i = 0.09 + 0.36 + 0.81 + 1.44 + 2.25 + 3.24 + 4.41 + 5.76 + 7.29 = 25.65

2 \sum_{i=1}^9 y_i = 2 \times 25.65 = 51.30

Substituting into the formula:

I = \frac{0.3}{2} [9.00 + 51.30] = 0.15 \times 60.30 = 9.045

This matches option D.

Final answer

Option (D)

Common trap

Multiplying the endpoints by 2 as well, rather than weighting only the interior points by 2.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

GATE questions are the work of the organising institutes. This site is independent and is not affiliated with them.