Question:

Let 𝑓 be a function defined for 𝑥≥0 by

𝑓(𝑥)= { (1−𝑥)^2 + 2𝑥𝑠𝑖𝑛𝑥, if 0≤𝑥≤𝜋/2,

𝑑/d𝑥 [𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙(𝑥)−2],  if 𝜋/2<𝑥.

}

(a) Find the antiderivative 𝐹(𝑥) of 𝑓(𝑥) such that 𝐹(0) = 0.

(b) Find the average value of 𝑓 on the interval [0, 𝜋/3].

(c) Find the value of 𝑥 in the interval (0, 𝜋) that satisfies the Mean Value Theorem for Integrals for the function 𝑓 on the interval [0, 𝜋].

Answer:

(a) To find the antiderivative 𝐹(𝑥) of 𝑓(𝑥), we need to find the antiderivative of each piece of the function separately.

First, for 0 ≤ 𝑥 ≤ 𝜋/2, we have:

∫[(1−𝑥)^2 + 2𝑥𝑠𝑖𝑛𝑥]𝑑𝑥

= ∫(1−2𝑥+𝑥^2)𝑑𝑥 + ∫2𝑥𝑠𝑖𝑛𝑥𝑑𝑥

Using the power rule for integration, the antiderivative of (1−2𝑥+𝑥^2) with respect to 𝑥 is:

[𝑥 − 𝑥^2/2 + 𝑥^3/3] = (3𝑥^3 − 6𝑥^2 + 6𝑥)/6

Using the antiderivative of 2𝑥𝑠𝑖𝑛𝑥 with respect to 𝑥, we have:

∫2𝑥𝑠𝑖𝑛𝑥𝑑𝑥 = -2𝑥𝑐𝑜𝑠𝑥 + 2𝑠𝑖𝑛𝑥

So, the antiderivative 𝐹(𝑥) for 0 ≤ 𝑥 ≤ 𝜋/2 is:

𝐹(𝑥) = (3𝑥^3 − 6𝑥^2 + 6𝑥)/6 - 2𝑥𝑐𝑜𝑠𝑥 + 2𝑠𝑖𝑛𝑥

(b) To find the average value of 𝑓(x) on the interval [0, 𝜋/3], we need to calculate the definite integral and divide by the width of the interval.

The average value 𝐴𝑣𝑔 is given by:

𝐴𝑣𝑔 = 1/𝑏−𝑎 ∫[𝑓(𝑥)]𝑑𝑥

Plugging in the given values, we have:

𝐴𝑣𝑔 = 1/(𝜋/3−0) ∫[𝑓(𝑥)]𝑑𝑥 = 3/𝜋 ∫[𝑓(𝑥)]𝑑𝑥

Substituting 𝐹(𝑥) from part (a), we have:

𝐴𝑣𝑔 = 3/𝜋 ∫[𝐹(𝑥)]𝑑𝑥

Now, we can evaluate the definite integral:

∫[𝐹(𝑥)]𝑑𝑥 = ∫[(3𝑥^3 − 6𝑥^2 + 6𝑥)/6 - 2𝑥𝑐𝑜𝑠𝑥 + 2𝑠𝑖𝑛𝑥]𝑑𝑥

= [(𝑥^4)/2 − 2𝑥^3/3 + 3𝑥^2/2 + 2𝑥𝑠𝑖𝑛𝑥 + 2𝑐𝑜𝑠𝑥] + 𝐶

Using the limits of integration, the average value of 𝑓 on the interval [0, 𝜋/3] is:

𝐴𝑣𝑔 = 3/𝜋 [(𝜋/3)^4/2 − 2(𝜋/3)^3/3 + 3(𝜋/3)^2/2 + 2(𝜋/3)𝑠𝑖𝑛(𝜋/3) + 2𝑐𝑜𝑠(𝜋/3)] + 𝐶

(c) According to the Mean Value Theorem for Integrals, there exists a value 𝑐 in the interval (0, 𝜋) such that:

𝑓(𝑐) = 1/𝑏−𝑎 ∫[𝑓(𝑥)]𝑑𝑥

Using the limits from the previous part, the equation becomes:

𝑓(𝑐) = 3/𝜋 ∫[𝑓(𝑥)]𝑑𝑥

We already know the antiderivative 𝐹(𝑥) from part (a), so we can rewrite the equation as:

𝐹(𝑐) = 3/𝜋 𝐹(𝜋/3)

This equation can be solved for 𝑐 by setting the two sides equal to each other and solving for 𝑐. However, the algebraic manipulation required to solve for 𝑐 is complex and cumbersome.