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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][tex]$F(x)=\int_6^x f(t) \, dt$[/tex][/tex]. Complete parts (a)-(c).

a. Evaluate [tex]A(4)[/tex] and [tex]A(5)[/tex]. Then use geometry to find an expression for [tex]A(x)[/tex] for all [tex]x \geq 3[/tex].

The value of [tex]A(4)[/tex] is [tex]\square[/tex]. (Simplify your answer.)

Sagot :

GinnyAnsweridn
Given the function [tex]\( f(t) = 4t - 12 \)[/tex], we need to evaluate the area function [tex]\( A(x) = \int_3^x f(t) \, dt \)[/tex].

### (a) Evaluation of [tex]\( A(4) \)[/tex] and [tex]\( A(5) \)[/tex]

1. Evaluate [tex]\( A(4) \)[/tex]:

[tex]\[ A(4) = \int_3^4 (4t - 12) \, dt \][/tex]

Let's find the antiderivative of [tex]\( f(t) \)[/tex]:

[tex]\[ \int (4t - 12) \, dt = 2t^2 - 12t \][/tex]

Now, evaluate this antiderivative from 3 to 4:

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

Calculate the values at the upper and lower limits:

Upper limit (when [tex]\( t = 4 \)[/tex]):

[tex]\[ 2(4)^2 - 12(4) = 2(16) - 48 = 32 - 48 = -16 \][/tex]

Lower limit (when [tex]\( t = 3 \)[/tex]):

[tex]\[ 2(3)^2 - 12(3) = 2(9) - 36 = 18 - 36 = -18 \][/tex]

Now, find [tex]\( A(4) \)[/tex]:

[tex]\[ A(4) = (-16) - (-18) = -16 + 18 = 2 \][/tex]

Thus, [tex]\( A(4) = 2 \)[/tex].

2. Evaluate [tex]\( A(5) \)[/tex]:

[tex]\[ A(5) = \int_3^5 (4t - 12) \, dt \][/tex]

Again, using the antiderivative [tex]\( 2t^2 - 12t \)[/tex]:

[tex]\[ A(5) = \left[ 2t^2 - 12t \right]_3^5 \][/tex]

Calculate the values at the upper and lower limits:

Upper limit (when [tex]\( t = 5 \)[/tex]):

[tex]\[ 2(5)^2 - 12(5) = 2(25) - 60 = 50 - 60 = -10 \][/tex]

Lower limit (when [tex]\( t = 3 \)[/tex]):

As before,

[tex]\[ 2(3)^2 - 12(3) = -18 \][/tex]

Now, find [tex]\( A(5) \)[/tex]:

[tex]\[ A(5) = (-10) - (-18) = -10 + 18 = 8 \][/tex]

Thus, [tex]\( A(5) = 8 \)[/tex].

3. Expression for [tex]\( A(x) \)[/tex] for all [tex]\( x \geq 3 \)[/tex]:

The general expression for [tex]\( A(x) \)[/tex] can be found by evaluating the definite integral from 3 to [tex]\( x \)[/tex]:

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

Using the antiderivative [tex]\( 2t^2 - 12t \)[/tex]:

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

Evaluate at the upper and lower limits:

Upper limit (when [tex]\( t = x \)[/tex]):

[tex]\[ 2x^2 - 12x \][/tex]

Lower limit (when [tex]\( t = 3 \)[/tex]):

[tex]\[ 2(3)^2 - 12(3) = 18 - 36 = -18 \][/tex]

So,

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

Thus, the expression for [tex]\( A(x) \)[/tex] for all [tex]\( x \geq 3 \)[/tex] is:

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

### Summary

- [tex]\( A(4) = 2 \)[/tex]
- [tex]\( A(5) = 8 \)[/tex]
- The expression for [tex]\( A(x) \)[/tex] for all [tex]\( x \geq 3 \)[/tex] is [tex]\( A(x) = 2x^2 - 12x + 18 \)[/tex].
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