Find the average value of the function defined by f(x)on the interval[ 0,pi^(1/4)]

Question

f(x) = x^3sin7x^4 on the interval [0,pi^(1/4)]

Answer 1

The average value of a function f over an interval [a, b] is given by:

[1 / (b - a)] * ∫_a^b f(x)dx

We will first find the indefinite integral of the function in the problem,

then take the necessary definite integral to find the average value of the function over the given interval.

Consider ∫ x³ sin(7x⁴) dx.

Let u = 7x⁴. Then

du = 7 * 4x³ dx = 28x³ dx

du = 28x³ dx

⇒ du / 28 = (28x³ dx) / 28

⇒ du / 28 = x³ dx

∫ x³ sin(7x⁴) dx

= ∫ sin(7x⁴) * x³ dx

= ∫ sin(u) * du / 28

= (1/28) ∫ sin(u) du

= (1/28) * [-cos(u)] + C

= - (1/28) cos (u) + C

= - (1/28) cos (7x⁴) + C

With the notation used at the beginning, in this problem, a = 0 and b = π^1/4.

So the average value of the function f over the interval [0, π^1/4] is given by:

[1 / (π^1/4 - 0)] * ∫₀^{π to the one fourth} x³ sin(7x⁴) dx

= (1 / π^1/4) * [- (1/28) cos (7x⁴)]_{x = 0}^{x = π to the one fourth}

= (1 / π^1/4) * {- (1/28) cos (7* (π^1/4)⁴) - [- (1/28) cos (7* (0^1/4)⁴)]}

= (1 / π^1/4) * [- (1/28) cos (7π) + (1/28) cos (0)]

= (1 / π^1/4) * [- (1/28) * (-1) + (1/28) * 1]

= (1 / π^1/4) * (1/28 + 1/28)

= (1 / π^1/4) * (1/14)

= 1 / (14π^1/4)

A few notes:

1. cos(0) = 1
2. cos(7π) = cos(π + 6π) = cos(π + 3*2π) = cos(π) = -1 because the cosine function is 2π periodic
3. 1/28 + 1/28 = 2/28 = 1 * 2 / (14 * 2) = 1 / 14
4. I wrote "π to the one fourth" in some places because this platform does not allow for a second superindex