Question

Use a computer algebra system and the error formulas to find \(n\) such that the error in the approximation of the definite integral is less than 0.00001 using (a) the Trapezoidal Rule and (b) Simpson's Rule. $$ \int_{0}^{1} \sin x^{2} d x $$

Short answer

For the Trapezoidal Rule, \(n\) must be greater than 257. For Simpson's Rule, \(n\) must be greater than 8.

Step-by-step solution

Trapezoidal Rule Error Formula

The error Formula can be expressed as \(E_{T}=K \frac{(b-a)^{3}}{12 n^{2}}\) where \(|f''(x)| \leq K\) for \(a \leq x \leq b\). Here, \(a = 0\), \(b = 1\) and \(K = 2\) as \(\left|\frac{{d^{2}}}{{dn^{2}}}\left(\sin x^{2}\right)\right| \leq 2\). Setting \(E_{T} < 0.00001\) and solving for \(n\) will give us the required value.

Solving for n (Trapezoidal Rule)

Substitute the given values into the formula and solve for \(n\). This gives us \(n > \sqrt{\frac{K(b-a)^{3}}{12E_{T}}}\). Substituting the known values gives \(n > \sqrt{\frac{2(1-0)^{3}}{12 \times 0.00001}}\). Solving for \(n\) gives \(n > 257.04\). This means that \(n\) must be an integer greater than 257 for the error to be less than 0.00001.

Simpson's Rule Error Formula

The error formula can be expressed as \(E_{S}=L \frac{(b-a)^{5}}{180 n^{4}}\) where \(|f''''(x)| \leq L\) for \(a \leq x \leq b\). Here, \(a = 0\), \(b = 1\) and \(L = 24\) as \(|f''''(x)| \leq 24\) for the given function. Setting \(E_{S} < 0.00001\) and solving for \(n\) will give us the required value.

Solving for n (Simpson's Rule)

Substitute the given values into the formula and solve for \(n\). This gives \(n > \sqrt[4]{\frac{L(b-a)^{5}}{180E_{S}}}\). Substituting the known values gives \(n > \sqrt[4]{\frac{24(1-0)^{5}}{180 \times 0.00001}}\). Solving for \(n\) gives \(n > 8.72\). This means that \(n\) must be an integer greater than 8 for the error to be less than 0.00001.