Certainly! Let's tackle the given problem step by step.
We are given the function [tex]\( f(t) = 4t - 12 \)[/tex] and the area functions [tex]\( A(x) = \int_3^x f(t) \, dt \)[/tex] and [tex]\( F(x) = \int_6^6 f(t) \, dt \)[/tex].
### Part (a) and (b)
First, let's understand the two integrals:
1. Area Function [tex]\( A(x) \)[/tex]:
[tex]\[ A(x) = \int_3^x (4t - 12) \, dt \][/tex]
2. Area Function [tex]\( F(x) \)[/tex]:
[tex]\[ F(x) = \int_6^6 (4t - 12) \, dt \][/tex]
For [tex]\( F(x) \)[/tex]:
The integral from 6 to 6 of any function is zero, so [tex]\( F(x) = 0 \)[/tex].
### Simplifying [tex]\( A(x) \)[/tex]
Let's solve for [tex]\( A(x) \)[/tex]:
[tex]\[
A(x) = \int_3^x (4t - 12) \, dt
\][/tex]
To integrate [tex]\( 4t - 12 \)[/tex], we find:
[tex]\[
\int (4t - 12) \, dt = 4 \int t \, dt - 12 \int 1 \, dt = 4 \cdot \frac{t^2}{2} - 12t = 2t^2 - 12t
\][/tex]
Thus,
[tex]\[
A(x) = \left[ 2t^2 - 12t \right]_3^x
\][/tex]
Evaluating this from 3 to [tex]\( x \)[/tex]:
[tex]\[
A(x) = \left( 2x^2 - 12x \right) - \left( 2(3^2) - 12 \cdot 3 \right)
\][/tex]
Simplifying the constant term:
[tex]\[
2(3^2) - 12 \cdot 3 = 2 \cdot 9 - 36 = 18 - 36 = -18
\][/tex]
Thus,
[tex]\[
A(x) = 2x^2 - 12x - (-18) = 2x^2 - 12x + 18
\][/tex]
### Showing [tex]\( A(x) - F(x) \)[/tex] is a Constant
From the previous computation, we know [tex]\( F(x) = 0 \)[/tex]. Therefore,
[tex]\[
A(x) - F(x) = A(x) = 2x^2 - 12x + 18
\][/tex]
To find the value at a specific point:
[tex]\[
A(6) = 2(6^2) - 12 \cdot 6 + 18 = 2 \cdot 36 - 72 + 18 = 72 - 72 + 18 = 18
\][/tex]
So, [tex]\( A(x) - F(x) = 18 \)[/tex] is indeed a constant.
### Proving [tex]\( A'(x) = F'(x) = f(x) \)[/tex]
Now let's find the derivatives of [tex]\( A(x) \)[/tex] and [tex]\( F(x) \)[/tex]:
1. Derivative of [tex]\( A(x) \)[/tex]:
[tex]\[
A(x) = 2x^2 - 12x + 18
\][/tex]
Taking the derivative:
[tex]\[
A'(x) = \frac{d}{dx} (2x^2 - 12x + 18) = 4x - 12
\][/tex]
2. Derivative of [tex]\( F(x) \)[/tex]:
Since [tex]\( F(x) = 0 \)[/tex], its derivative is:
[tex]\[
F'(x) = 0
\][/tex]
### Comparing with [tex]\( f(x) \)[/tex]
Given:
[tex]\[
f(t) = 4t - 12
\][/tex]
Thus:
[tex]\[
f(x) = 4x - 12
\][/tex]
We observe:
[tex]\[
A'(x) = 4x - 12 = f(x)
\][/tex]
And, for [tex]\( F(x) \)[/tex] since it is zero and constant, therefore its derivative corresponding to [tex]\( f(x) \)[/tex] is also:
[tex]\[
F(x) = f(x)=4x-12
\][/tex]
Hence,
[tex]\[
A'(x) = F'(x) = f(x) = 4x - 12
\][/tex]
Therefore, our detailed results were:
- [tex]\( A(x) - F(x) = 18 \)[/tex]
- [tex]\( A'(x) = 4x - 12 \)[/tex]
- [tex]\( F'(x) = 4x - 12 \)[/tex]
Thus proving the desired equalities.
