Question

area

Answer 1

the bounds here would be -4.71097 and 0

Answer 2

then it would be the integral of tan(x)^-1 - (x^2 + 5x) from -4.71 to 0

Answer 3

i think

Answer 4

So to me it seems like

\[\Large x(x+5) < y<\tan^{-1}(x) \\\]
And
\[ \Large-4.71097 < x < 0 \]

Answer 5

The area bounded below by arctan x and above by y=x^2 + 5x is on the first quadrant.

Answer 6

isn't it the 3rd quadrant?

Answer 7

http://i.imgur.com/9eAza.png

Answer 8

Did you write the problem correctly?

Answer 9

yes

Answer 10

\[\Large \int_{-4.71}^0\int_{x(x+5)}^{\tan^-1(x} dydx\]

Answer 11

http://www.wolframalpha.com/input/?i=plot+y%3Dtan%5E-1%28x%29+and+y%3Dx%5E2+%2B+5x Did I copy it wrong?

Answer 12

that is correct

Answer 13

The area in the third quadrant is bounded above by arctan x and below by y= x^2 + 5 x

Answer 14

but your integral doesnt make sense

Answer 15

it is a double integral @daniellerner, if you write it in the iterated form you obtain the same answer.

Answer 16

it is below arctan x and above x^2 + 5x look at the raph

Answer 17

how would you do the double integral

Answer 18

I am arguing about above and below.

Answer 19

Are you sure you have the above and below written correctly?

Answer 20

http://www.wolframalpha.com/input/?i=arctan%28x%29+%3D+x%5E2+%2B+5x
tan(x)^-1 is the purple one

Answer 21

and that is above

Answer 22

No, x^2 + 5 x is the purple one.

Answer 23

however my integral simplifies to, if you don't want to do it in the multivariable form

\[\Large \int_{-4.71}^0 (\tan^{-1}(x)-x(x+5))dx \]

Answer 24

http://www.wolframalpha.com/input/?i=plot++y%3Dx^2+%2B+5x

Answer 25

Bye.

Answer 26

Why are you leaving? What you are saying doesn't make sense..

Answer 27

spacelimbus do you know what that simplifies to

Answer 28

i got 15.79

Answer 29

I haven't calculated the integral yet, you can check your answer with wolframalpha though if you care about the numerical answer.

It's not the easiest integral because you have to apply integration by parts.

Answer 30

i just plugged it into my ti89

Answer 31

I also believe that a big misunderstanding in this question was going on at all time, sir eliassaab was right when he mentioned his point, I believe it was a confusion about how the question is written.

Answer 32

hmm well if that's a legit way at your school then do that (-: Otherwise integrate piecewise and by parts.

Answer 33

yeah this is the calculator part of the hw

Answer 34

ok let me feed wolf with it and check.

Answer 35

ok thanks

Answer 36

Yes you got it right:

http://www.wolframalpha.com/input/?i=integrate+from+-4.71+to+0+%28tan%5E%28-1%29x-x%28x%2B5%29%29

Answer 37

Nice!

Answer 38

well i have another question hopefully you can help

Answer 39

You can post it, I recommend you opening a new separate question for it though, so in case I can't, I am sure others will.

Answer 40

yeah i posted as a new one