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).
Antiderivative (one of) of y is
4sin(x/4)(x^2-32)+32x*cos(x/4)
Sorrr, I didn't ask what to find. I have to Find the formula for F(a).
To find the area between the x-axis and the graph of y = x^2cos(x/4) between x = 0 and x = a, we need to evaluate the definite integral of the function between those limits.
The integral can be written as follows:
F(a) = ∫[0 to a] (x^2*cos(x/4)) dx
To evaluate this integral, we need to find its antiderivative first. The antiderivative of a function is also known as the indefinite integral, which represents the family of functions whose derivative is the original function.
To find the antiderivative of x^2*cos(x/4), we can use integration techniques such as integration by parts or a substitution. For this example, let's use a substitution.
Let u = x/4, then du = (1/4) dx.
Rearranging the equation, we have dx = 4 du.
Now substituting these values in our integral:
F(a) = ∫[0 to a] (x^2*cos(x/4)) dx
= ∫[0 to a] (16*u^2*cos(u)) du
= 16 * ∫[0 to a] (u^2*cos(u)) du
With this substitution in place, we can now integrate the function u^2*cos(u) with respect to u. The integral of cos(u) is sin(u), and the integral of u^2 is (1/3)*u^3.
Applying these evaluations to the integral:
F(a) = 16 * [ (1/3)*u^3*sin(u) - ∫[(0 to a) (1/3)*u^3*sin(u)) ] du
The remaining integral can be evaluated using integration techniques such as integration by parts or a substitution.
Once the integral is evaluated, substitute the limit a into the equation F(a) to get the final value of the area between the x-axis and the graph of y = x^2*cos(x/4) between x = 0 and x = a.
It is important to note that due to the complexity of the integral and the need for further calculations, finding an explicit formula for F(a) may not be feasible. In such cases, numerical methods such as numerical integration or approximation methods can be used to estimate the area.