Correct answer is (d) 0.35
Easy explanation: Given: \(∫_0^8x^{\frac{1}{3}}dx, n = 4,\)
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