Calculate the integral approximations of T8 & M8 for the integral of 4 to 9 of x^2 dx . Answer must be accurate to 8 decimal places .

Question

anonymous · Answer

What do you mean by T8 and M8?

anonymous · Answer

It's T sub 8 & M sub 8 . I'm really not so sure because I have another question that is S sub 8 .

dumbcow · Answer

$$\int\limits_{4}^{9}x^{2} dx = \frac{1}{3}x^{3}\left(\begin{matrix}9 \ 4\end{matrix}\right) = \frac{9^{3}}{3}-\frac{4^{3}}{3}$$

anonymous · Answer

T = trapezoidal & M = midpoint .

anonymous · Answer

Alright, so you mean to find the approximate value of $\int_4^9 x^2 dx $. Both of those methods involve in estimating the area under the graph of $f(x)=x^2$ with quadrangles. I'm not totally familiar with your notation but I'd say the number 8 means there to divide the area to 8 subsections.

anonymous · Answer

Yes , but how do I have 8 subsections for 4 - 9 when there's only 5 ?

dumbcow · Answer

each subsection will have to be (5/8)

dumbcow · Answer

i forget how to do trapezoid..lol
for midpoint, you will have to find the following
f(5/16), f(15/16), f(25/16) ...
then multiply each one by (5/8)
then add them all up

anonymous · Answer

Ka-ching! http://en.wikipedia.org/wiki/Trapezoidal_rule