Question

Given \(∫_0^8x^\frac{1}{3}dx,\) find the error in approximating the integral using Simpson’s 1/3 Rule with n=4.

(a) 1.8

(b) 2.9

(c) 0.3

(d) 0.35

I have been asked this question during an online exam.

My enquiry is from Change of Order of Integration: Double Integral topic in division Multiple Integrals of Engineering Mathematics

Answer 1

The correct answer is (d) 0.35

To elaborate: Given: \(∫_0^8x^\frac{1}{3}dx,n = 3,\)

Let \(f(x)= x^\frac{1}{3},\)

\(∆x = \frac{b-a}{2}= \frac{8-0}{2}=4\) ………………since b=8, a=0 (limits of the given integral)

Hence endpoints xi have coordinates {0, 2, 4, 6, 8}.

Calculating the function values at xi, we get,

\(f(0)= 0^\frac{1}{3}=0\)

\(f(2)= 2^\frac{1}{3}\)

\(f(4)= 4^\frac{1}{3}\)

\(f(6)= 6^\frac{1}{3}\)

\(f(8)= 8^\frac{1}{3}  =2\)

Substituting these values in the formula,

\(∫_0^8x^\frac{1}{3}  dx ≈ \frac{∆x}{3}  [f(0)+4f(2)+2f(4)+4f(6)+f(8)]\)

 \(    ≈\frac{2}{3}[0+4(2^\frac{1}{3})+2(4^\frac{1}{3})+ 4(6^\frac{1}{3})+2]  ≈ 11.65\)

Actual integral value,

\(∫_0^8x^\frac{1}{3} dx= \left(\frac{x^\frac{4}{3}}{\frac{4}{3}}\right)_0^8=12\)

Error in approximating the integral = 12 – 11.65 = 0.35