Question

Use the trapezium rule with three intervals to find an approximation to:

Answer 1

@518nad

Answer 2

@zepdrix

Answer 3

do u know any languages?

Answer 4

Oh, sorry, I didn't put it up.

Answer 5

\[\int\limits_{0}^{?12} \sqrt{4+x^2} dx\]

Answer 6

how many intervals

Answer 7

Three

Answer 8

Intervals should be 0, 4, 8, 12?

Answer 9

those are the points for your intervals

Answer 10

so your intervals are
(0,4)
(4,8)
(8,12)

Answer 11

Yes, that's what I mean. Thank you.

Answer 12

just making sure

Answer 13

ok carry on

Answer 14

rad(4+0^2) + rad(4+4^2) * 2 = area for the first one?

Answer 15

Yes, the 2 should be multiplying the entire thing though,
2 * ( rad(4+0^2) + rad(4+4^2) )

Answer 16

That's what I meant, sorry.

Answer 17

Soo, I got 79.2 for the final answer when rounded to 3 sig figs.
Area(first interval) 12.9443, Area(2nd) 25.4367, Area(3rd) 40.8235

Answer 18

Integrating will give you 77.977.
So yes, your calculations are probably correct \c:/

Answer 19

Oh, there's another part to the question, find: \[\int\limits_{?}^{?} 4\cos(\frac{ 1 }{ 3 }x + 2)dx\]
Is this supposed to be Integration?

Answer 20

It has no limits.

Answer 21

There is this rule (that I made up) ...
I like to call it the Linear Coefficient Rule:

If \(\large\rm \frac{d}{dx}F(x)=f(x)\),
then,\[\large\rm \int\limits F(ax)=\frac1a f(ax)\]

Answer 22

This only works with linear coefficients.
If the coefficient is on say, an x^2, then something else is going on.
But as long as the x is linear, first degree,
we can simply divide by the coefficient when we integrate.

Answer 23

Hopefully this makes a little bit of sense if you think back to your chain rule.
When you differentiate,\[\large\rm \frac{d}{dx}\sin(8x)\quad=\quad \cos(8x)\cdot 8\]You get an extra 8 `multiplying` because of the inner function.

When you integrate, you LOSE that 8.
So you actually end up `dividing it out`.

Answer 24

\[\large\rm \int\limits \cos(8x)dx=\frac18\sin(8x)\]

Answer 25

Ok I rambled on for a bit there...
Do you understand why this helps though?

It allows us to avoid using a substitution.

Answer 26

It makes sense, yes. Even with your reasoning, it still equals 12sin(1/3x +2)dx, at least that's what I see from what you showed.

Answer 27

AHH I wrote my integral backwards, woops.\[\large\rm if~\frac{d}{dx}F(x)=f(x)\]
\[\large\rm then~\int\limits f(ax)=\frac1a F(ax)\]Not a big deal though >.<

Answer 28

So you divided a 1/3 out,
which is the same as multiplying by 3?

Yayyy good job

Answer 29

Oh, yeah, don't you have to add + C?

Answer 30

Yes yes, that would be a good idea :)

Answer 31

Alright wonderful! Thank you so much. I have another question, but it's different. Well, I have two, but I think I can do one on my own, but the other one seems to look confusing. I will tag you?

Answer 32

I gotta go for a lil bit.
You can tag and maybe I'll come by if I find time.

Danny boy super smart too though :D

Answer 33

Alrighty. That's fine, thank you again!