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Can anyone help me with solving this integral?

Mathematics
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\[\int\limits_{2}^{5} (4-2x)dx\]
which part is giving you night terrors?
lol the 2nd one!

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Other answers:

what is the integration rule for x^n ?
\[\int\limits_{2}^{5} 4.dx - 2\int\limits_{2}^{5}x.dx\]
or you can think of it in terms of derivatives; what is the derivative rule for x^n ?
nx^n-1
good, now how would you undo that? which is all an integration is ... undoing a derivative
thats wht i am having trouble with!
divide by n and add 1 back to the exponent
\[x=x^1\]right?
x^1
yea
\[x^1=x^{2-1}\]so n=2 in this case
ohh yea right!
so, does x^2 derive down to 2x?
yea it will be 2x^2-1
but we need x!
but we already have a 2x
you dont HAVE to pull out the constant. In this case it acutally helps out
2x/2
\[\frac d{dx}x^2=2x\] \[\int 2x~dx=x^2\]
ohh gotch yea so, its gona be like (5-2)^2
\[\int\limits_{2}^{5} 4.dx - \int\limits_{2}^{5}2x.dx\] \[4x(5,2) - x^2(5,2)\] \[[4(5)-4(2)] - [(5)^2-(2)^2]\]
so i was doing just 1 term of x
you were mixing up the subtraction and the function into an ungodly abomination :)
yea right, i got it now! Thank you very much.. :)
\[\int_{a}^{b} f(x)~dx=F(b)-F(a)~NOT~F(b-a)\]
-9
20-8-(25-4) 20-8-25+4 24 - 33 = -9 yes
yayy.. thanks again!

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