Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
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.
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.
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
