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Let [tex]f(t) = 4t - 12[/tex] and consider the two area functions [tex]A(x) = \int_3^x f(t) \, dt[/tex] and [tex]F(x) = \int_6^x f(t) \, dt[/tex].

Complete parts (a)–(c).

(a) The value of [tex]F(8)[/tex] is 32. (Simplify your answer.)

(b) Use geometry to find an expression for [tex]F(x)[/tex] when [tex]x \geq 6[/tex].

[tex]F(x) = 2x^2 - 12x[/tex]

(Type an expression using [tex]x[/tex] as the variable.)

(c) Show that [tex]A(x) - F(x)[/tex] is a constant and that [tex]A'(x) = F'(x) = f(x)[/tex].

First, prove [tex]A(x) - F(x)[/tex] is a constant. Using the expressions for [tex]A(x)[/tex] and [tex]F(x)[/tex] found in previous steps, subtract [tex]F(x)[/tex] from [tex]A(x)[/tex].

The value of [tex]A(x) - F(x)[/tex] is [tex]\square[/tex]. (Simplify your answer.)

Sagot :

GinnyAnswer
Let's solve the problem step-by-step.

### Part (a)
The value of [tex]\( F(8) \)[/tex]

To find [tex]\( F(8) \)[/tex], we integrate the function [tex]\( f(t) = 4t - 12 \)[/tex] from 6 to 8:
[tex]\[ F(x) = \int_6^x (4t - 12) \, dt \][/tex]

Evaluating [tex]\( F(8) \)[/tex]:
[tex]\[ F(8) = \int_6^8 (4t - 12) \, dt \][/tex]

The result is:
[tex]\[ F(8) = 32 \][/tex]
Thus, the value of [tex]\( F(8) \)[/tex] is 32.

### Part (b)
Expression for [tex]\( F(x) \)[/tex] when [tex]\( x \geq 6 \)[/tex]

To find [tex]\( F(x) \)[/tex], we need to integrate [tex]\( f(t) = 4t - 12 \)[/tex] from 6 to [tex]\( x \)[/tex]:
[tex]\[ F(x) = \int_6^x (4t - 12) \, dt \][/tex]

[tex]\[ F(x) = \left[ 2t^2 - 12t \right]_6^x \][/tex]
[tex]\[ F(x) = \left( 2x^2 - 12x \right) - \left( 2(6)^2 - 12(6) \right) \][/tex]

[tex]\[ F(x) = 2x^2 - 12x - (72 - 72) \][/tex]
[tex]\[ F(x) = 2x^2 - 12x \][/tex]

So, [tex]\( F(x) = 2x^2 - 12x \)[/tex] when [tex]\( x \geq 6 \)[/tex].

### Part (c)
Show that [tex]\( A(x) - F(x) \)[/tex] is a constant and that [tex]\( A'(x) = F'(x) = f(x) \)[/tex]

First, let's find [tex]\( A(x) \)[/tex]:

[tex]\[ A(x) = \int_3^x (4t - 12) \, dt \][/tex]
[tex]\[ A(x) = \left[ 2t^2 - 12t \right]_3^x \][/tex]

[tex]\[ A(x) = \left( 2x^2 - 12x \right) - \left( 2(3)^2 - 12(3) \right) \][/tex]
[tex]\[ A(x) = 2x^2 - 12x - (18 - 36) \][/tex]
[tex]\[ A(x) = 2x^2 - 12x + 18 \][/tex]

Now, subtract [tex]\( F(x) \)[/tex] from [tex]\( A(x) \)[/tex]:
[tex]\[ A(x) - F(x) = (2x^2 - 12x + 18) - (2x^2 - 12x) \][/tex]
[tex]\[ A(x) - F(x) = 18 \][/tex]

Therefore, [tex]\( A(x) - F(x) \)[/tex] is a constant, specifically [tex]\( 18 \)[/tex].

Next, let's verify that [tex]\( A'(x) = F'(x) = f(x) \)[/tex]:

[tex]\[ A(x) = 2x^2 - 12x + 18 \][/tex]
[tex]\[ A'(x) = \frac{d}{dx} (2x^2 - 12x + 18) = 4x - 12 \][/tex]

[tex]\[ F(x) = 2x^2 - 12x \][/tex]
[tex]\[ F'(x) = \frac{d}{dx} (2x^2 - 12x) = 4x - 12 \][/tex]

We see that:
[tex]\[ A'(x) = 4x - 12 \][/tex]
[tex]\[ F'(x) = 4x - 12 \][/tex]
And:
[tex]\[ f(x) = 4x - 12 \][/tex]

Thus, we have shown that [tex]\( A'(x) = F'(x) = f(x) \)[/tex].

So, the value of [tex]\( A(x) - F(x) \)[/tex] is 18.
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