Integral calculus is a cornerstone of many competitive exams, including JEE Mathematics. While analytical integration is crucial, understanding and implementing numerical integration methods is vital for solving problems where analytical solutions are difficult or impossible to obtain. This module focuses on implementing a numerical integration method, specifically the Trapezoidal Rule, as a practical approach to approximating definite integrals.

Analytical integration involves finding an antiderivative. However, many functions do not have elementary antiderivatives (e.g., $e^{-x^2}$), or the antiderivative might be too complex to find. Numerical integration provides a way to approximate the value of a definite integral by dividing the area under the curve into smaller, manageable shapes whose areas can be easily calculated.

The formula for the Trapezoidal Rule with $n$ subintervals is:

$\int_{a}^{b} f(x) dx \approx \frac{h}{2} [f(x_0) + 2f(x_1) + 2f(x_2) + \dots + 2f(x_{n-1}) + f(x_n)]$

where $h = \frac{b-a}{n}$, and $x_i = a + ih$ for $i = 0, 1, \dots, n$.

To implement this, you'll need to:

1. Define the function $f(x)$ you want to integrate.

2. Specify the interval of integration $[a, b]$.

3. Choose the number of subintervals, $n$.

4. Calculate the width of each subinterval, $h = \frac{b-a}{n}$.

5. Calculate the sum according to the formula, summing $f(x_0)$ and $f(x_n)$ once, and $f(x_i)$ for $i=1, \dots, n-1$ twice.

Let's approximate $\int_{0}^{1} x^2 dx$ using the Trapezoidal Rule with $n=4$.

Here, $a=0$, $b=1$, $f(x) = x^2$, and $n=4$. The width $h = \frac{1-0}{4} = 0.25$.

The points are $x_0=0, x_1=0.25, x_2=0.5, x_3=0.75, x_4=1$.

The function values are:

$f(x_0) = f(0) = 0^2 = 0$

$f(x_1) = f(0.25) = (0.25)^2 = 0.0625$

$f(x_2) = f(0.5) = (0.5)^2 = 0.25$

$f(x_3) = f(0.75) = (0.75)^2 = 0.5625$

$f(x_4) = f(1) = 1^2 = 1$

Applying the formula:

$\int_{0}^{1} x^2 dx \approx \frac{0.25}{2} [f(0) + 2f(0.25) + 2f(0.5) + 2f(0.75) + f(1)]$

$\approx 0.125 [0 + 2(0.0625) + 2(0.25) + 2(0.5625) + 1]$

$\approx 0.125 [0 + 0.125 + 0.5 + 1.125 + 1]$

$\approx 0.125 [2.75] = 0.34375$

The exact value of $\int_{0}^{1} x^2 dx$ is $\left[\frac{x^3}{3}\right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \approx 0.33333$. The Trapezoidal Rule with $n=4$ gives a good approximation.

The accuracy of the Trapezoidal Rule (and other numerical integration methods) generally improves as the number of subintervals ($n$) increases. For competitive exams, you might be asked to use a specific value of $n$ or to determine how many subintervals are needed to achieve a certain level of accuracy.

While the Trapezoidal Rule is a good starting point, other methods offer even better accuracy for the same number of subintervals. These include:

Simpson's Rule, which uses parabolic segments to approximate the curve, is often preferred for its superior accuracy and is frequently tested in competitive exams.