- #1

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so Work = the integral from [0,9] of 10/(1+x)^2(dx) U substitution gives

u=1+x

du=dx

the 10 is factored out tilll the end.

Ok so, in this example at x=0, U=1 and at x=9, U=10

Now the new limits are [1,10] instead if [0,9]. But Why?

- Thread starter bmed90
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- #1

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so Work = the integral from [0,9] of 10/(1+x)^2(dx) U substitution gives

u=1+x

du=dx

the 10 is factored out tilll the end.

Ok so, in this example at x=0, U=1 and at x=9, U=10

Now the new limits are [1,10] instead if [0,9]. But Why?

- #2

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- #3

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Can you elaborate more on this? Get basic if you have to. If you don't want to that is also ok.

- #4

HallsofIvy

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[tex]\int_{x= 0}^9 \frac{dx}{(1+ x)^2}[/tex]

are

[tex]\int_{u= 1}^{10} \frac{du}{u^2}[/tex]

(It's not a bad idea to actually write the "x= " or "u= " in the integral like that. Especially when you get to double and triple integrals.)

Notice that this is -1/u so evaluating betweeen 1 and 10 gives -1/10+ 1= 9/10.

If you go back to "x" the integral is -1/(x+ 1) which you evaluate between 0 and 9: -1/(9+ 1)- (-1/(0+1)= -1/10+ 1= 9/10 again.

- #5

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If you plot y = 1+x on an xy-plane, it is just a line that starts at x=1 with slope 1. Take note of what the area looks like under the plot on [0,9].

Then plot v = u on a uv-plane, which is just a line that starts at u=0 with slope 1 (it's the above graph shifted down). If we find the area under this new graph on [0,9], it will clearly be less than the area on the original graph. If we instead calculate the area on [1,10], it will then be the same. The good news is that the function u = 1+x tells us exactly how to change the limits of integration.

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