dessyj1
  • dessyj1
NUmber 8
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Hi!! What is the Number 8 for?
misty1212
  • misty1212
\[\infty\]
dessyj1
  • dessyj1
\[if \left(\begin{matrix}8-x^2 for -2\le x \le2 \\ x^2 else where,\end{matrix}\right) then \int\limits_{-1}^{3}f(x) dx is a number \between \]

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dessyj1
  • dessyj1
is a number between?* would you like to know the choices?
dessyj1
  • dessyj1
how would i do this?
dessyj1
  • dessyj1
ignore number 8
xapproachesinfinity
  • xapproachesinfinity
hmm i see peiewize function and you the area under its curve
dessyj1
  • dessyj1
it is a peicewise
xapproachesinfinity
  • xapproachesinfinity
what 8 do yu want us to ignore? 8-x^2? this should be part of the funtion
dessyj1
  • dessyj1
i am talking about the title of the question. that is what you should ignore
xapproachesinfinity
  • xapproachesinfinity
eh soka!
misty1212
  • misty1212
\[f(x) = \left\{\begin{array}{rcc} 8-x^2 & \text{if} & -2\leq x\leq 2\\ x^2& \text{otherwise} & \end{array} \right. \] just seeing if i could do it, ignore me
xapproachesinfinity
  • xapproachesinfinity
alright you just do some linearity here see where you should integrate 8-x^2 and x^2
xapproachesinfinity
  • xapproachesinfinity
as you can see from -1
xapproachesinfinity
  • xapproachesinfinity
the rest 2 to 3 you integrate x^2
xapproachesinfinity
  • xapproachesinfinity
do you get it, or should i write it down?
dessyj1
  • dessyj1
i know the answer is 8
dessyj1
  • dessyj1
The question asks for what the answer could be between
xapproachesinfinity
  • xapproachesinfinity
\[\int_{-1}^{3}f(x)dx=\int_{-1}^{2}f(x)dx+\int_{2}^{3}f(x)dx\] \[\int_{-1}^{2}(8-x^2)dx+\int_{2}^{3}x^2dx\]
dessyj1
  • dessyj1
The choices were a) 0 and 8 b) 8 and 16 c) 16 and 24 d) 24 and 32 e) 32 and 40
xapproachesinfinity
  • xapproachesinfinity
hmm i see so they separated it you just do those two integral separately
xapproachesinfinity
  • xapproachesinfinity
\[\int_{-1}^{2}(8-x^2)dx=?\]
dessyj1
  • dessyj1
43/3
xapproachesinfinity
  • xapproachesinfinity
really how ?
dessyj1
  • dessyj1
the integral is 8x-(1/3)x^3
xapproachesinfinity
  • xapproachesinfinity
i found 62/3
dessyj1
  • dessyj1
i tried it again, it gave me 21
dessyj1
  • dessyj1
i made an arithmetic mistake.
xapproachesinfinity
  • xapproachesinfinity
yes 21 is good!
xapproachesinfinity
  • xapproachesinfinity
but then it is not in the choices
dessyj1
  • dessyj1
I believe the answer is B
dessyj1
  • dessyj1
Earlier,I said the solution to the intergral was 8 and i was wrong
xapproachesinfinity
  • xapproachesinfinity
something is wrong in what you wrote? that way must work
xapproachesinfinity
  • xapproachesinfinity
can you post a snap shot of the problem please
xapproachesinfinity
  • xapproachesinfinity
is the function exately like missy wrote it
xapproachesinfinity
  • xapproachesinfinity
hey yo there?
xapproachesinfinity
  • xapproachesinfinity
???
xapproachesinfinity
  • xapproachesinfinity
i'm waiting...
dessyj1
  • dessyj1
sorry, i will do that now
dessyj1
  • dessyj1
1 Attachment
xapproachesinfinity
  • xapproachesinfinity
now i got what is those btw mean we need to add those two integrals and see the solution lies btw what two points
xapproachesinfinity
  • xapproachesinfinity
i found 21 +19/3 =82/3 which lies 24<82/3<32
xapproachesinfinity
  • xapproachesinfinity
go with what we found in the first time and add the other integral
xapproachesinfinity
  • xapproachesinfinity
see if you got same as me
xapproachesinfinity
  • xapproachesinfinity
just miss understood the problem
dessyj1
  • dessyj1
i got 21-(19/3)
xapproachesinfinity
  • xapproachesinfinity
hmm why -19/3?
dessyj1
  • dessyj1
never mind
dessyj1
  • dessyj1
we are good.
dessyj1
  • dessyj1
so D?
xapproachesinfinity
  • xapproachesinfinity
\[\int_{2}^{3}x^2ds=\frac{x^3}{3}=19/3\]
xapproachesinfinity
  • xapproachesinfinity
yes D
dessyj1
  • dessyj1
Instead of getting the sum of both integrals, I decided to do the difference.
xapproachesinfinity
  • xapproachesinfinity
yeah i see that :)
xapproachesinfinity
  • xapproachesinfinity
alright i think we solved that :)
dessyj1
  • dessyj1
Arigato!
xapproachesinfinity
  • xapproachesinfinity
No problem :) yakoso

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