1. ## Simple integration problem

Okay so I know how to integrate really well, but I came across a problem that left me confused.

x[0,1] f(x) = (1-x)^9 dx

I was told that when integrating this problem to put a negative sign in front of the integrand, but not why. Why would you do this?

Okay so I know how to integrate really well, but I came across a problem that left me confused.

x[0,1] f(x) = (1-x)^9 dx

I was told that when integrating this problem to put a negative sign in front of the integrand, but not why. Why would you do this?
Try the substitution u= 1- x. Then $(1- x)^9$ becomes $u^9$ which is easy to integrate. But you also have to replace "dx". With u= 1- x, du/dx= -1 so du= -dx and dx= -du. Your integral becomes $-\int u^9(-du)= -\int u^9 du$.

Notice that I didn't put the limits of integration here. You could integrate $-u^9$, to get $-1/10 u^{10}$, of course, change that to $-1/10 (1-x)^9$ and evaluate that at x=0 and x=1.

But I prefer to change the limits of integration as I substitute. Those are x= 0 and x= 1 and they have to be changed to u also. When x= 0, u= 1 and when x= 1, u= 0.
$\int_0^1 (1-x)^9 dx= -\int_1^0 u^9 du$
That "-" in "1- x" has reversed the order of the limits of integration- but $\int_a^b f(x)dx= -\int_b^a f(x)dx$ so you really have
$\int_0^1 (1-x)^9 dx= -\int_1^0 u^9 du= \int_0^1 u^9 du$
and the negative has disappeared!

Another way of looking at it is to remember that integration is the opposite of differentiation. Whatever function you integrated to, you would have f(u)= f(1- x). And you would use the "chain rule" to differentiate: the derivative of f(1-x) is the derivative of f times the derivative of 1- x, which is -1.

With any substitution for a decreasing function, like 1- x, the derivative will be negative. That's why you need the "-" in the anti-derivative.

3. oh wow I get it.