Please help?
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).
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.
To find the formula for F(a), we need to evaluate the definite integral of the function y = x^2cos(x/4) with respect to x, from x = 0 to x = a.
First, let's find the antiderivative (also called the indefinite integral) of the function x^2cos(x/4). To do this, we can use the method of integration by parts, combined with the trigonometric identity cos(x/4) = (1/2)(cos(x/4 + x/2) + cos(x/4 - x/2)):
∫(x^2cos(x/4)) dx = x^2 * (1/2)(cos(x/4 + x/2) + cos(x/4 - x/2)) dx.
Now, we can apply the integration by parts formula, which states that ∫u dv = uv - ∫v du, with u = x^2 and dv = (1/2)(cos(x/4 + x/2) + cos(x/4 - x/2)) dx.
Differentiating u, we have du = 2x dx, and integrating dv, we obtain:
v = (1/2)∫(cos(x/4 + x/2) + cos(x/4 - x/2)) dx.
Now, let's evaluate the integral v using the sum and difference formulas for cosine:
v = (1/2)∫(cos(x/4)cos(x/2) - sin(x/4)sin(x/2)) dx.
Using the substitution u = x/2, we can rewrite the integral as:
v = (1/2)∫(cos(x/4)cos(u) - sin(x/4)sin(u)) 2 du.
Simplifying, we get:
v = ∫(cos(x/4)cos(u) - sin(x/4)sin(u)) du.
Applying the sum and difference formulas for cosine, we have:
v = ∫(cos(x/4)cos(u) - sin(x/4)sin(u)) du = ∫(cos(u + x/4)) du.
Using the integral of cosine, we find:
v = sin(u + x/4) + C.
Now, substituting back u = x/2, we have:
v = sin(x/2 + x/4) + C.
Finally, going back to the original integral, we get:
∫(x^2cos(x/4)) dx = x^2 * (1/2)(sin(x/2 + x/4) + C).
Plugging in the limits of integration, x = 0 and x = a, we can find the formula for F(a):
F(a) = ∫(0 to a) (x^2cos(x/4)) dx = (a^2 / 2)(sin(a/2 + a/4)) - (0^2 / 2)(sin(0/2 + 0/4)).
Simplifying further, we have:
F(a) = (a^2 / 2)(sin(5a/4)).
Therefore, the formula for F(a) is given by (a^2 / 2)(sin(5a/4)).