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).
To find the formula for F(a), we need to find the integral of the function y = x^2cos(x/4) with respect to x over the interval [0, a].
The integral of x^2cos(x/4) with respect to x is not a standard integral, so we will need to use integration techniques to evaluate it.
The first step is to apply integration by parts. Let's define u = x^2 and dv = cos(x/4) dx. Taking the derivative of u with respect to x gives du = 2x dx, and integrating dv gives v = 4sin(x/4).
Using the integration by parts formula, ∫u dv = uv - ∫v du, we have:
∫x^2cos(x/4) dx = x^2 * 4sin(x/4) - ∫(4sin(x/4)) * (2x dx)
Simplifying further, we get:
∫x^2cos(x/4) dx = 4x^2sin(x/4) - 8∫xsin(x/4) dx
Now, we need to apply integration by parts again to evaluate the remaining integral. Let u = x, dv = sin(x/4) dx. Taking the derivative of u gives du = dx, and integrating dv gives v = -4cos(x/4).
Using the integration by parts formula again, we have:
∫xsin(x/4) dx = x * (-4cos(x/4)) - ∫(-4cos(x/4)) dx
Simplifying further, we get:
∫xsin(x/4) dx = -4xcos(x/4) + 16∫cos(x/4) dx
Now, the integral of cos(x/4) is a standard integral and evaluates to 4sin(x/4). Substituting this back into the equation, we have:
∫xsin(x/4) dx = -4xcos(x/4) + 16(4sin(x/4)) + C
where C is the constant of integration.
Substituting this result back into the earlier equation, we have:
∫x^2cos(x/4) dx = 4x^2sin(x/4) - 8(-4xcos(x/4) + 16(4sin(x/4))) + C
Simplifying further, we get:
∫x^2cos(x/4) dx = 4x^2sin(x/4) + 32xcos(x/4) - 256sin(x/4) + C
Therefore, the formula for F(a), the area between the x-axis and the graph of y = x^2cos(x/4) between x = 0 and x = a, is:
F(a) = 4a^2sin(a/4) + 32acos(a/4) - 256sin(a/4) + C
where C is the constant of integration.