Use composite simpsons rule Answer problem # 2 /2cosh(x sin 0 )de.I Jo2L(x) = =Use the composite Simpson's 1/3 rule with step length h = T/16 to estimate I, (3.5).Answer. 7.378202Use the composite Simpson's 1/3 rule with step length h = 0.4 to estimateIn(2 +3x)dx,1+x2I = Composite Simpson's rule [ edit ]If the interval of integration [a, b) is in some sense "small", then Simpson's rule with n = 2 subintervals will provide an adequate approximation to the exact integral. By "small" we mean that the function beingintegrated is relatively smooth over the interval [a, b]. For such a function, a smooth quadratic interpolant like the one used in Simpson's rule will give good results.However, it is often the case that the function we are trying to integrate is not smooth over the interval. Typically, this means that either the function is highly oscillatory or lacks derivatives at certain points. Inthese cases, Simpson's rule may give very poor results. One common way of handling this problem is by breaking up the interval [a, b] into n 2 small subintervals. Simpson's rule is then applied to eachsubinterval, with the results being summed to produce an approximation for the integral over the entire interval. This sort of approach is termed the composite Simpson's rule.Suppose that the interval a, b is split up into n sub-intervals, with n an even number. Then, the composite Simpson's rule is given byn/2f(x) dx =E [f(®2j_2) + 4f(x2j–1) + f(x2;)]j=1п/2-1n/2f(x0)+2 f(x2;) + 4 f(x2j–1) + f(xn)3j=1j=1where x; = a + jh for j = 0, 1,..., n – 1, n with h =a)/n; in particular, xo = a and xn = b. This composite rule with n = 2 corresponds with the regular Simpson's Rule of the preceding section.The error committed by the composite Simpson's rule ish4(b –180a) f(4) (£),where { is some number between a and b, and h =(b – a)/n is the "step length".3] The error is bounded (in absolute value) byh4()b180a) max |f(4 (£)|-fe(a,b]-

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Use composite simpsons rule Answer problem # 2 /2cosh(x sin 0 )de.I Jo2L(x) = =Use the composite Simpson's 1/3 rule with step length h = T/16 to estimate I, (3.5).Answer. 7.378202Use the composite Simpson's 1/3 rule with step length h = 0.4 to estimateIn(2 +3x)dx,1+x2I = Composite Simpson's rule [ edit ]If the interval of integration [a, b) is in some sense &#34;small&#34;, then Simpson's rule with n = 2 subintervals will provide an adequate approximation to the exact integral. By &#34;small&#34; we mean that the function beingintegrated is relatively smooth over the interval [a, b]. For such a function, a smooth quadratic interpolant like the one used in Simpson's rule will give good results.However, it is often the case that the function we are trying to integrate is not smooth over the interval. Typically, this means that either the function is highly oscillatory or lacks derivatives at certain points. Inthese cases, Simpson's rule may give very poor results. One common way of handling this problem is by breaking up the interval [a, b] into n > 2 small subintervals. Simpson's rule is then applied to eachsubinterval, with the results being summed to produce an approximation for the integral over the entire interval. This sort of approach is termed the composite Simpson's rule.Suppose that the interval a, b is split up into n sub-intervals, with n an even number. Then, the composite Simpson's rule is given byn/2f(x) dx =E [f(®2j_2) + 4f(x2j–1) + f(x2;)]j=1п/2-1n/2f(x0)+2 f(x2;) + 4 f(x2j–1) + f(xn)3j=1j=1where x; = a + jh for j = 0, 1,..., n – 1, n with h =a)/n; in particular, xo = a and xn = b. This composite rule with n = 2 corresponds with the regular Simpson's Rule of the preceding section.The error committed by the composite Simpson's rule ish4(b –180a) f(4) (£),where { is some number between a and b, and h =(b – a)/n is the &#34;step length&#34;.3] The error is bounded (in absolute value) byh4()b180a) max |f(4 (£)|-fe(a,b]-

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Use the composite Simpson's 1/3 rule with step length h = 0.4 to estimate In(2 + 3/x) dx. 1+x² I =