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integrate ∫(4-x)^11 dx u = 4-x ∫(4-x)^11 dx = ∫u^11 du = u^12 / 12 + C rite?
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Yes, now sub back in
I think.. >.>
= (4-x)^12 / 12 + C
i looked at the solution it says that it is: - ∫u^11 du why is that?
what's du?
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You're wrong shamil.. u=4-x du=-dx
The integral becomes: -Int(u^11du)
me no comprende, why though? o.o
u = 4-x right?
yeh
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u = 4x^0 - x^1 du/dx = 0*4x^(-1) - 1*x^0
du/dx = -1
Screw it, bah's got it.
oh
du=-dx
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ty bah
= -u^12 / 12 + C = -(4-x)^12/12 + C
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