Question

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

(a) 1.8

(b) 2.9

(c) 0.3

(d) 0.35

This question was posed to me in examination.

I need to ask this question from Derivatives of Area topic in section Differential Calculus of Engineering Mathematics

Answer 1

Correct answer is (d) 0.35

Easiest explanation: Given: \(∫_0^8 x^{\frac{1}{3}} dx,\),  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