Use the form of the definition of the integral given in this theorem to evaluate the integral. 0 to 1 of (x^3 − 4x^2) dx

Question

Use the form of the definition of the integral given in this theorem to evaluate the integral. 0 to 1 of (x^3 − 4x^2) dx

Answer 1

Divide the interval [0,1] into n equal subintervals each of length Δx = (1-0)/n = 1/n

The right endpoints of the intervals are 1/n, 2/n, 3/n, 4/n, n/n. So, the right endpoint of interval i is

x_i = i/n.

Let f(x) = x³ - 4x²

The value of the given integral is lim_n→∞∑_{i = 1 to n} [(i/n)³ - 4(i/n)²](1/n)

= lim_n→∞ [(1/n⁴)∑_{i=1 to n}(i)³ - 4(1/n³)∑_{i =1 to n} (i)²]

= lim_n→∞ [(1/n⁴)n²(n+1)²/4 - (4/n³)n(n+1)(2n+1)/6]

= lim_{n→ ∞} [(n⁴+2n³+n²) / (4n⁴) - (8n³+12n²+4n) / (6n³)]

= lim_n→∞ [1/4 + 1/(2n) + 1/(4n²) - 4/3 - 2/n - 2/(3n²)] = 1/4 - 4/3 = -13/12

Answer 2

∫₀¹ x³ − 4x² dx To evaluate integrals for polynomial, the exponent increases by 1 and also needs to be put into the denominator. Each term can be done separately.

x⁴/4− 4x³/3 now, evaluate between 0 and 1

[(1)⁴/4− 4(1)³/3] - [(0)⁴/4− 4(0)³/3]

[1/4 - 4/3] - [0]

[3/12 - 16/12] = -13/12