Definite Integral of an Area function

Question

Let A(x) = ∫^x₀ 3cos(t) dt.

1. Evaluate A(𝛑)
2. Describe what A(𝛑) means
3. Compute A'(x)

Answer 1

By the Fundamental Theorem of calculus,

the function A(x) is one of the antiderivatives of 3cos(t).

All the antiderivatives of 3cos(t) are of the form 3sin(t)+C, for some constant C.

The definite integral is then

A(x) = (3sin(x)+C) - (3sin(0)+C) = 3sin(x)

1. A(π) = 3sin(π) = 0

2. A(π) is the area under 3cos(t) on the interval [0,π].

It it 0 because on the interval [0,π/2] 3cos(t) is positive, and

on the interval [π/2, π] 3cos(t) is negative.

The negative area on [π/2,π] is as large as the positive area on [0,π/2].

3.A'(x) = 3cos(x)