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I had to compute $$T_2$$ using the trapezoid rule formula:

$$ T_m (f) = \frac{h}{2}(f(a) + f(b)) + h \sum\limits_{i=1}^{2^m -1} f(x_i) $$

Values are the following:

[a, b] = [0, 2]

m = 2

h = 1/2

f(x) = x^5 + x

f(0) = 0
f(1) = 2
f(2) = 34

From my calculations, the value of the summation should be

$$ \frac{1}{2} [ 6(3) + 34(3)] $$

but that is incorrect and I believe the answer should be something like :

$$ \frac{1}{2} [11.7] $$

I am unable to evaluate the summation properly. A step by step example of what computations occur within the summation sign would be really helpful.

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  • $\begingroup$ Yup, sorry. Edited it @Moo $\endgroup$ – x89 May 10 '20 at 19:40
  • $\begingroup$ What you computed is $T_0$. Where does the $(3)$ come from, what does it mean? With the 3 values given you could compute $T_1$. $T_2$ is computed from a subdivision with $2^m=4$ intervals. $\endgroup$ – Lutz Lehmann May 11 '20 at 5:36
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We are given

$$\begin{align}f(x) &= x^5 + x\\ [a, b] &= [0, 2]\\ m &= 2\\ h &= \dfrac{1}{2} \end{align}$$

For $n = 2$, we have

$$ T_2 (f) = \dfrac{1}{2}(f(a) + 2 f(x_1)+ f(b)) $$

We get

$$T_2(f) = \dfrac{1}{2}( 0 + 2(2) + 34) = \dfrac{1}{2}(38) = 19$$

The actual value is

$$\int_0^2 (x^5+x)~dx = \dfrac{38}{3} \approx 12.6667$$

If you use three sub-intervals

$$T_3(x) = 15.5637860082305$$

If we use ten sub-intervals

$$T_{10}(x) = 12.9328$$

Update

Using your formula, we have

$$T_m (f) = \frac{h}{2}(f(a) + f(b)) + h \sum\limits_{i=1}^{2^m -1} f(x_i) = \dfrac{1}{4} (f(a) + f(b)) + \dfrac{1}{2} (f(x_1) + f(x_2) + f(x_3))$$

This means we have $x_1 = 1/2, x_2 = 1, x_3 = 3/2$.

This gives

$\dfrac{1}{4} (f(0) + f(2))+ \dfrac{1}{2} (f(1/2) + f(1) + f(3/2))\\ = \dfrac{1}{4}(0 + 34) + \dfrac{1}{2}(17/32 + 2 + 291/32) \\= 229/16 \approx 14.3125$

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  • $\begingroup$ How did you compute T2? Why are you using 2f(x1)? Could you please explain how you derived that? My idea was that 1/4 * (0 x 34) + summation $\endgroup$ – x89 May 11 '20 at 0:13
  • $\begingroup$ I think the value of T2 should be around 14. According to the sample answer I had. $\endgroup$ – x89 May 11 '20 at 0:17
  • $\begingroup$ I am using the formula I gave in the qs (1st eq) $\endgroup$ – x89 May 11 '20 at 1:22
  • $\begingroup$ Yup, it's h, not h/2. I checked online $\endgroup$ – x89 May 11 '20 at 12:55
  • $\begingroup$ google.com/… $\endgroup$ – x89 May 11 '20 at 12:57

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