anonymous
  • anonymous
someone help? (:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\int\limits_{-2}^{0} (\frac{ 1 }{ 2 }t^4+\frac{ 1 }{ 4 }t^3-t)dt\]
nubeer
  • nubeer
Hint. \[\int\limits_{?}^{?} t ^{4} =\frac{ t ^{4+1} }{ 4+1}\]
anonymous
  • anonymous
I honestly don't even know how to solve or begin to solve for it.. can you help me step by step? I have 7 more problems like this and I'd like to do them all myself.. But I need help solving and the steps to do so.

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More answers

anonymous
  • anonymous
are you familiar with the general formula of integration?
anonymous
  • anonymous
I'm familiar with... \[\int\limits_{a}^{b} f(x)dx=F(x)|_{a}^{b} = F(b)-F(a)\]
nubeer
  • nubeer
ya this is done when we apply the limits but do you know how we do integration?
anonymous
  • anonymous
I don't.
anonymous
  • anonymous
the general formula for integration of \(\int x^n dx = \frac{x^n+1}{n+1} +C\)
nubeer
  • nubeer
|dw:1355140748339:dw|
anonymous
  • anonymous
|dw:1355140890023:dw|
anonymous
  • anonymous
@winterfez i believe that's \(\left[ \frac{1}{2} \frac{t^5}{5}+\frac{1}{4} \frac{t^4}{4}- \frac{t^2}{2} \right]^0_{-2}\)
anonymous
  • anonymous
\[(\frac{ 1 }{ 2 }\frac{ 0 }{ 5 }+\frac{ 1 }{ 4 }\frac{ 0^{4} }{ 4 }-\frac{ 0 }{ 2 })-(\frac{ 1 }{ 2 }\frac{ (-2)^5 }{ 5 }+\frac{ 1 }{ 4 }\frac{ (-2)^4 }{ 4 }-\frac{ (-2)^{2} }{ 2 })\]
anonymous
  • anonymous
now use your calculator to punch in those suckers
anonymous
  • anonymous
can you explain the [...]_{-2}^{0}
anonymous
  • anonymous
∫abf(x)dx=F(x)|ba=F(b)−F(a) <--this is why.
anonymous
  • anonymous
from your original post up there.
anonymous
  • anonymous
ohhh so i do apply that.. I just missed a step.
anonymous
  • anonymous
yup :) i think you've got it.
anonymous
  • anonymous
thank you guys! (: you're amazing!
anonymous
  • anonymous
you're welcome :)
anonymous
  • anonymous
I got 21/5.. does that seem right @Shadowys
anonymous
  • anonymous
-21/5?
anonymous
  • anonymous
err... let me recheck my work.
anonymous
  • anonymous
sorry, my mistake. i typed wrong. lol no neg.
anonymous
  • anonymous
oh okay (: is that the answer though? because what happens with the .... +C?
anonymous
  • anonymous
well. that's for indefinite integrals because you don't know where is the limit. the C is gone when it is definite.
anonymous
  • anonymous
oh okay (: so 21/5 it is?
anonymous
  • anonymous
yup! :)
anonymous
  • anonymous
thanks again!
anonymous
  • anonymous
you're welcome :)

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