monroe17
someone help? (:
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monroe17
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\[\int\limits_{-2}^{0} (\frac{ 1 }{ 2 }t^4+\frac{ 1 }{ 4 }t^3-t)dt\]
nubeer
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Hint.
\[\int\limits_{?}^{?} t ^{4} =\frac{ t ^{4+1} }{ 4+1}\]
monroe17
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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.
Shadowys
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are you familiar with the general formula of integration?
monroe17
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I'm familiar with... \[\int\limits_{a}^{b} f(x)dx=F(x)|_{a}^{b} = F(b)-F(a)\]
nubeer
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ya this is done when we apply the limits but do you know how we do integration?
monroe17
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I don't.
Shadowys
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the general formula for integration of \(\int x^n dx = \frac{x^n+1}{n+1} +C\)
nubeer
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|dw:1355140748339:dw|
winterfez
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|dw:1355140890023:dw|
Shadowys
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@winterfez i believe that's \(\left[ \frac{1}{2} \frac{t^5}{5}+\frac{1}{4} \frac{t^4}{4}- \frac{t^2}{2} \right]^0_{-2}\)
winterfez
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\[(\frac{ 1 }{ 2 }\frac{ 0 }{ 5 }+\frac{ 1 }{ 4 }\frac{ 0^{4} }{ 4 }-\frac{ 0 }{ 2 })-(\frac{ 1 }{ 2 }\frac{ (-2)^5 }{ 5 }+\frac{ 1 }{ 4 }\frac{ (-2)^4 }{ 4 }-\frac{ (-2)^{2} }{ 2 })\]
winterfez
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now use your calculator to punch in those suckers
monroe17
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can you explain the [...]_{-2}^{0}
Shadowys
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∫abf(x)dx=F(x)|ba=F(b)−F(a) <--this is why.
Shadowys
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from your original post up there.
monroe17
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ohhh so i do apply that.. I just missed a step.
Shadowys
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yup :) i think you've got it.
monroe17
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thank you guys! (: you're amazing!
Shadowys
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you're welcome :)
monroe17
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I got 21/5.. does that seem right @Shadowys
Shadowys
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-21/5?
monroe17
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err... let me recheck my work.
Shadowys
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sorry, my mistake. i typed wrong. lol no neg.
monroe17
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oh okay (: is that the answer though? because what happens with the .... +C?
Shadowys
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well. that's for indefinite integrals because you don't know where is the limit. the C is gone when it is definite.
monroe17
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oh okay (: so 21/5 it is?
Shadowys
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yup! :)
monroe17
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thanks again!
Shadowys
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you're welcome :)