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ok i need huge help on this since the roots is 31...http://www.webassign.net/cgi-bin/symimage.cgi?expr=int_0%2A%2A4%20%28x%20-%202%29%2A%2A%2831%29%20dx
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u=x-2 du= dx \[\int u^{31} du\] from 2 to 6
no 4-0
4 to 0
I know since u= x-2 , we change the limits of integration
how?
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where 6 comes from?
its 2 to -2
yes sorry it is -2 to 2
ok so now what i do?
u^31 -> 1/32 u^32 evaluate from -2 to 2
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so this woild be: 1/32 (x-2)^32 correct? then i plug in the limits of 2 to -2.....(1/32 (2-2)^32)-(1/32(-2-2)^32=
0-1/32(-4)^32 right?
if you are going to change it back to x, leave that limit as they were 0 to 4
no im using the 2 to -2 as the limit..
so what would that be?
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\[\frac{1}{32}u^{32}\] \[\frac{1}{32}(2^{32}-(-2)^{32})\]
thats the answer?
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