- #1
nickadams
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Say we want to find ∫(-cosx)sinx dx
set u = -cosx
so du/dx = sinx
∫u [itex]\frac{du}{dx}[/itex]dxAnd then the dx's apparently cancel out? What is going on? I thought [itex]\frac{du}{dx}[/itex] meant lim Δx→0 [itex]\frac{u(x+Δx)-u(x)}{Δx}[/itex] and that the dx in the context of an integral was a representation of the Δx's from the lim n→∞ [itex]\sum^{n}_{i=1}[/itex]f([itex]x_{i^{*}}[/itex])Δx?
So how is this canceling justified with the above def'ns?
set u = -cosx
so du/dx = sinx
∫u [itex]\frac{du}{dx}[/itex]dxAnd then the dx's apparently cancel out? What is going on? I thought [itex]\frac{du}{dx}[/itex] meant lim Δx→0 [itex]\frac{u(x+Δx)-u(x)}{Δx}[/itex] and that the dx in the context of an integral was a representation of the Δx's from the lim n→∞ [itex]\sum^{n}_{i=1}[/itex]f([itex]x_{i^{*}}[/itex])Δx?
So how is this canceling justified with the above def'ns?
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