Approximating Definite Integrals — Question 7

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Question 7

Show that Simpson’s Rule with n=2n=2 is exact for ∫02x3dx.\int_0^2x^3\,dx.

Original worksheet page 1: question and worked solution for 1-10-007
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Question 7 – Solution

Step 1: Find the width and nodes. h=2−02=1,x0=0,x1=1,x2=2.h=\frac{2-0}{2}=1,\qquad x_0=0,\ x_1=1,\ x_2=2. Step 2: Apply Simpson’s Rule. S2=h3[f(x0)+4f(x1)+f(x2)]=13[03+4(13)+23]=13(0+4+8)=4.\begin{align*} S_2&=\frac h3[f(x_0)+4f(x_1)+f(x_2)]\\ &=\frac13[0^3+4(1^3)+2^3]\\ &=\frac13(0+4+8)=4. \end{align*} Step 3: Evaluate the exact integral. I=∫02x3dx=[x44]02=164−0=4.\begin{align*} I&=\int_0^2x^3\,dx =\left[\frac{x^4}{4}\right]_0^2\\ &=\frac{16}{4}-0=4. \end{align*} Since S2=IS_2=I, Simpson’s Rule is exact here. S2=I=4\boxed{S_2=I=4}

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