Evaluates as the limit of sums: ∫(x^2 - 1) dx for x ∈ [1,2]

Question

Evaluates as the limit of sums: \(\int\limits_1^2 (x^2 - 1) dx\)

Answer 1

Let f(x) = x² – 1 for 1 ≤ x ≤ 2 

We divide the interval [1, 2] into n equal sub-intervals each of length h

We have a = 1, b = 2

Here, f(a) = f(1) = (1)² – 1 = 0 

f(a + h) = f(1 + h) = (1 + h)² – 1 = 1 + h² + 2h – 1 = h² + 2h 

f(a + 2h) = f(1 + 2h) = (1 + 2h)² – 1 = 1 + 4h² + 4h – 1 = 4h² + 4h 

f[a + (n - 1)h] = f[1 + (n – 1)h] = [1 + (n – 1)h]² – 1 = 1 + (n – 1)² h² + 2(n – 1 )h – 1 = (n – 1)² h² + 2 (n – 1 )h