We can calculate the value of a definite integral by using trapezoids to divide the area under the curve for the given function.
Trapezoidal Rule Statement: Let f(x) be a continuous function on the interval (a, b). Now divide the intervals (a, b) into n equal sub-intervals with each of width,
Δx = (b – a)/n, such that a = x0 < x1 < x2 < x3 <…..< xn = b
Then the Trapezoidal Rule formula for area approximating the definite integral b∫af(x)dx is given by:
b∫af(x) dx ≈ Tn = △x/2 [f(x0) + 2f(x1) + 2f(x2) +….2f(xn-1) + f(xn)]
where, xi = a + i△x
If n → ∞, R.H.S of the expression approaches the definite integral b∫a f(x)dx
Proof:
To prove the trapezoidal rule, consider a curve as shown in the figure above and divide the area under that curve into trapezoids. We see that the first trapezoid has a height Δx and parallel bases of length y0 or f(x0) and y1 or f1. Thus, the area of the first trapezoid in the above figure can be given as,
(1/2) Δx [f(x0) + f(x1)]
The areas of the remaining trapezoids are (1/2)Δx [f(x1) + f(x2)], (1/2)Δx [f(x2) + f(x3)], and so on.
Consequently,
∫ba f(x) dx ≈ (1/2)Δx (f(x0)+f(x1) ) + (1/2)Δx (f(x1)+f(x2) ) + (1/2)Δx (f(x2)+f(x3) ) + … + (1/2)Δx (f(n-1) + f(xn) )
After taking out a common factor of (1/2)Δx and combining like terms, we have,
∫ba f(x) dx≈ (Δx/2) (f(x0)+2 f(x1)+2 f(x2)+2 f(x3)+ … +2f(n-1) + f(xn) )
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