# Changing the Limits of Integration

1. Mar 24, 2010

### bmed90

Ive seen some example of U substitution where the limits of integration are changed for example, Say we have a particle pushed along the x axis with force=10/(1+x)^2 and we want the work required to move it 9 ft.

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. Mar 24, 2010

### pbandjay

You have limits x=0 to x=9. When you substitute u = 1 + x, you no longer integrate with respect to x. You integrate with respect to u, so you must make sure to change the limits to values of u, instead of x. Luckily you have a nice formula for u. When x=0, u = 1+0 = 1, when x=9, u = 1+9 = 10. So you now want to integrate from u=1 to u=10. Does this help?

3. Mar 25, 2010

### bmed90

no longer integrate with respect to x --> You integrate with respect to u, so you must make sure to change the limits to values of u

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

4. Mar 25, 2010

### HallsofIvy

The limits of integration in
$$\int_{x= 0}^9 \frac{dx}{(1+ x)^2}$$
are x values. When you make the substituion u= 1+ x, you must change everything from "x" to "u". When x= 0, u= 1+ 0= 1 and when x= 9, u= 1+ 9= 10:

$$\int_{u= 1}^{10} \frac{du}{u^2}$$

(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. Mar 25, 2010

### pbandjay

Maybe it would also help to draw some graphs that represent the change of variables.

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.