This is from James Stewart Calculus Textbook When Is Simpson's Rule Exact?(a) Show that if f is a polynomial of degree 3 or lower, thenSimpson's Rule gives the exact value of f(x) dx.(b) Find the approximation S4 for f(x³ - 6x² + 4x) dx andverify that S4 is the exact value of the integral.(c) Use the error bound given in (4) to explain why thestatement in part (a) must be true.

Given: f is a polynomial of degree less than equal to 3. ∫08x3-6x2+4xdx To do: Show that for such f, Simpson's rule gives the exact value of ∫abfxdx. Find the approximation S4 for the given integral and show that S4 is the exact value of the integral. Using the error bound of Simpson's rule to say that a is true.(a) Let fx be of the form fx=ax3+bx2+cx+d The degree of fx is less than or equal to 3 depends upon the values of the coefficients. We shall prove that value of the integral in two sub intervals is equal to the Simpson's rule. Let node points of the two sub interval be xi-1, xi, xi+1, where ∆x=h=b-an. We shall integrate fx over the interval xi-1,xi+1. Consider, ∫xi-1xi+1ax3+bx2+cx+ddx Let t=x-xi ⇒ x=t+xi ⇒dx=dt When x=xi-1, t=xi-1-xi=xi-1-xi-1-h=h Also, when x=xi+1, t=xi+1-xi=xi+h-xi=h So, the integral will be ∫xi-1xi+1ax3+bx2+cx+ddx=∫-hhat+xi3+bt+xi2+ct+xi+ddtThe above integral can be written as ∫-hhat+xi3+bt+xi2+ct+xi+ddt=∫-hhAt3+Bt2+Ct+Ddt where, A=aB=3axi+bC=3axi2+2bxi+cD=axi3+bxi2+cxi+d Let the new function be gt i.e., gt=At3+Bt2+Ct+D Since t3 and t are odd functions on the interval -h,h the integral becomes zero. Also, t2 is an even function. Then, the integral can be further simplified as follows. ∫-hhAt3+Bt2+Ct+Ddt=∫-hhAt3dt+∫-hhBt2dt+∫-hhCtdt+∫-hhDdt=A0+B∫-hht2dt+C0+D∫-hhdt=2B∫0ht2dt+D∫-hhdt=2Bt330h+Dt-hh=2Bh33-0+2Dh=2Bh33+2Dh=h32Bh2+6D=h3g-h+4g0+gh ∵ g-h+4g0+gh=2Bh2+6D Further, ∫-hhAt3+Bt2+Ct+Ddt=h3g-h+4g0+gh=h3fxi-1+4fxi+fxi+1 which is the formula for Simpson's rule for two sub intervals. Hence, for a polynomial of degree less than or equal to 3, the Simpson's rule gives the exact value of the integral.(b) Consider the given integral, ∫08x3-6x2+4xdx For n=4, Simpson's rule is ∫abfxdx=∆x3fx0+4fx1+2fx2+4fx3+fx4 Here, a=0, b=8 and ∆x=8-04=2. So, the node points are x0=0, x1=2, x2=4, x3=6, x4=8 Then, the integral can be evaluated as follows ∫abfxdx=∆x3fx0+4fx1+2fx2+4fx3+fx4=23f0+4f2+2f4+4f6+f8=230+-32+-32+96+160=23-32-32+96+160=23-64+96+160=23192=128Exact value of the integral can be calculated as follows. ∫08x3-6x2+4xdx=x44-6x33+4x2208=844-283+282=40964-2512+264=1024-1024+128=128 Note that both the values i.e., value using Simpson's rule and exact value are same which is equal to 128. Hence proved.(c) In evaluating an integral of the form ∫abfxdx, the error bound for the Simpson's rule is given by Es≤Kb-a5180n4 where, f4x≤K for x∈a,b Given that f is a polynomial of degree less than equal to 3. So, the 4th derivative of f vanishes. i.e., f4x=0 for all x. This implies, K=0. Then, the error bound becomes Es≤Kb-a5180n4=0b-a5180n4=0 From this we can conclude that there is no error in computing the integral value ∫abfxdx using Simpson's rule. Therefore, Simpson's rule gives the exact value for a polynomial of degree less than or equal to 3. Hence proved.