I am a bit confused as to why a problem in my book is using A(t) instead of D(t) in teh setup for simpsons rule. Why is the integral at the end setup like:$$\int_0^43200 A(t)dt$$ and not $$\int_0^43200 D(t)dt$$Here are the passages:...Read more

I have a question that is supposed to be very easy in a test:We approximate $\displaystyle \int_0^1 x^2$ with Simpsonrule, and 5 intervals. Choose solution:Firstly I don't understand what is meant with 5 intervals. Are the intervals not supposed to be even?Secondly, it's an easy test question. How can we get the reply fast without calculating$\frac{h}{3}(y_0+4y_1+y_2)?$...Read more

Let I denote the integral $I = \int_0^{\pi/2}\sqrt{\sin x}dx$ and 4 strips.Give a function for which Simpson’s rule returns an exact value.I just entered in the exact values (so $\sin(\pi/2)$ for example) into the formula but did not get an exact value I don't understand what this question wants and how to approach it....Read more

The trapezoidal rule applied on $ \int_0^{2} [f(x)] dx$ gives the value 5 and the Midpoint rule gives the value 4. What value does Simpson's rule give?So we have that T=f(0)+f(2).The Simpson's value is S=(1/3)*(f(0) + 4f(1)+ f(2))f(1) here is equal to 4 since its the midpoint value..I dont how how to combine these together to find the simpson's value...Read more

In this problem we will approximate the integral of $\cos(x^3)$ over the interval $[0, 2]$.(a) Write an expression for MN, TN and SN with $N = 4$.(b) For each of the approximations determine an N so that the error is guaranteedto be less than $10^{-4}.$ You will find it useful to know that on the interval $[0, 2], |\frac {d^4\cos(x^3)}{dx^4} | \le 850$.Please help me with the question. Thanks a lot!...Read more

Let n be even. Show how the composite Simpson rule with 2n equally spaced nodes can be computed from the case of n equally spaced nodes with a minimum amount of additional work.I've been working on this problem for almost 2 hours now and can't seem to get anywhere. I feel it has something to do with the initial conditions on the [a,b] interval ($x_i=a+ih \text{ and } h=\frac{b-a}{2}$) and working the problem so there are the same number of terms to calculate in 2n as there are in n. There's always replacing any instance of n with 2n and halving...Read more

They claim later on in the proof that $G’(0)=0$. I don’t see this. I would say that we have$$G(t)=\int_0^tF(\tau)d\tau-\int_0^{-t}F(\tau)d\tau-t/3[F(-t)+4F(0)+F(t)],$$which yields$$G’(0)=F(0)-F(0)-1/3[6\cdot F(0)]=-2F(0).$$Could someone explain what is going on?...Read more

They claim later on in the proof that $G’(0)=0$. I don’t see this. I would say that we have$$G(t)=\int_0^tF(\tau)d\tau-\int_0^{-t}F(\tau)d\tau-t/3[F(-t)+4F(0)+F(t)],$$which yields$$G’(0)=F(0)-F(0)-1/3[6\cdot F(0)]=-2F(0).$$Could someone explain what is going on?...Read more

In this problem we will approximate the integral of $\cos(x^3)$ over the interval $[0, 2]$.(a) Write an expression for MN, TN and SN with $N = 4$.(b) For each of the approximations determine an N so that the error is guaranteedto be less than $10^{-4}.$ You will find it useful to know that on the interval $[0, 2], |\frac {d^4\cos(x^3)}{dx^4} | \le 850$.Please help me with the question. Thanks a lot!...Read more