Question

Integration help. question below. will give medal

Answer 1

\[\int\limits_{0}^{3}\frac{ 1 }{ 16+x ^{2} } dx\]

I ended up with 9.2175 as my answer, but in the question, it said to show that the integral of the above sum is equal to 0.1609

Answer 2

I've already done part a and got my answer.

Answer 3

Mmm ok, so Trig Substitution, yes?

Answer 4

You ended up with 9 something for part A? D:
You may have done something wrong there... an approximation method should get you CLOSE TO the correct value.

Answer 5

\[\Large\rm x=4\tan \theta\]Have you dealt with trig sub before?

Answer 6

Using trig substitution, I got
\[\frac{ 1 }{ 4 }\tan^{-1} (\frac{ x }{ 4 })\]

Answer 7

Or rather, using the standard integral of the function

Answer 8

Mmm ok yah that seems correct.
Evaluating it at the given boundary values,\[\Large\rm \frac{1}{4}\left[\arctan\left(\frac{3}{4}\right)-\arctan\left(0\right)\right]\approx 0.1609\]

Answer 9

Oh, what I did incorrectly was separate it into two different functions. My bad. Thank you! :)

Answer 10

Oh, you silly billy :O

Answer 11

AHHHH, problem again. Typing this into my calculator gave me 9 point something :( @zepdrix

Answer 12

arctan(0) is just zero.
Try to remember that so you don't have to punch it into the calculator.

Answer 13

\[\Large\rm \tan^{-1}\left(\frac{3}{4}\right)\approx 0.6435\]Are you able to get this correct value from the calculator?

Answer 14

No, I actually got 36.8699

Answer 15

Should my calculator be set in radian mode or?

Answer 16

Ah yes.
You're getting \(\Large\rm 36.8699^o\)

Answer 17

Yes
Let me change it to radian mode and see

Answer 18

I'm getting it now in radian mode :D Thank you!

Answer 19

yay!