crammy90 said:
integrating
-30(20-t) as function of t
how do we know whether to do
a) (expand) = -600 + t
then integrate = -600t + (t^2)/2
or
b) -30*([20-t)^2)/2
= -15([20-t)^2)
which is doing the a^(n+1) all 0ver (n+1)
Well...I would pretty much expand all the time, especially if its a nice small expansion like this.
=∫-30(20-t) dt
= ∫-600 +
30t dt
= -600t +15t<sup>2</sup> + C
∫x<sup>n</sup> dx = x<sup>n+1</sup>/n+1
Only works for powers of your variable.
(If you do three unit, or four unit, you can make a substitution, where you change the variable of integration to do a question like this. )
For example, consider:
∫(x+1)<sup>2</sup> dx
By expanding,
=∫x<sup>2</sup> + 2x +1 dx
=x<sup>3</sup>/3 + x<sup>2</sup> + x + C
Trying to use the other method:
=∫(x+1)<sup>2</sup> dx
=(x+1)<sup>3</sup>/3 + C
The two methods are clearly unequal, and since we know the first method is correct, the second method must be incorrect.
