Please help?

Question

Let F(a) be the area between the x-axis and the graph of y=x^2cos(x/4) between x=0 and x=a, for a>0 (consider the area to be negative if the graph lies below the axis).

Find the formula for F(a).

Answer 1

Since we are considering negative areas, all we have to do is simple integration from 0 to a, without worrying about absolute value, etc.

So, the question is just how to integrate x^2 cos(x/4). The answer is: use integration by parts. Huh? How's that work?

Remember the product rule for doing derivatives?

(uv)' = u'v + uv'

If we rearrange terms a bit, we get

uv' = (uv)' - u'v
u * dv = d(u*v) - du * v

Now integrate:

Int(u dv) = Int(d(uv)) - Int(v du)
Int(u dv) = uv - Int(v du)

So, thr teick is to break up uv into two functions, one of which gets simpler as we take derivatives, and the other which doesn't get too complicated as we integrate.

In this problem, the logical choice is

u = x^2 because when we differentiate, the power goes down each time
dv = cos(x/4) because we can integrate that with no trouble.

So, now we set up our little chart:

u = x^2
du = 2x dx

dv = cos(x/4)
v = 4 sin(x/4)

Int(x^2 cos(x/4) dx) = x^2 * 4 sin(x/4) - Int(2x * 4 sin(x/4))
= 4 x^2 sin(x/4) - 8 Int(x sin(x/4))

Now we do it all over again, to get rid of the lingering x:

u = x
du = dx

dv = sin(x/4)
v = -4 cos(x/4)

Int(x sin(x/4)) = -4x cos(x/4) - Int(-4 cos(x/4) dx)
= -4x cos(x/4) + 16 sin(x/4)

Collect terms, and you have your final, somewhat messy answer.