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Use the graph of f(t) = 2t + 4 on the interval [-4, 6] to write the function F(x), where F(x)=the integral from x to 2 of f(t)dt
A F(x) = x2 + 6x
B F(x) = x2 + 4x - 12
C F(x) = x2 + 4x - 8
D F(x) = x2 + 8x - 20

Respuesta :

LammettHash
When [tex]x<2[/tex], you have

[tex]F(x)=\displaystyle\int_x^2(2t+4)\,\mathrm dt[/tex]
[tex]F(x)=(t^2+4t)\bigg|_{t=x}^{t=2}[/tex]
[tex]F(x)=(2^2+4(2))-(x^2+4x)[/tex]
[tex]F(x)=12-4x-x^2[/tex]

Even when [tex]x>2[/tex], you would get the same answer.

[tex]F(x)=\displaystyle\int_x^2(2t+4)\,\mathrm dt=-\int_2^x(2t+4)\,\mathrm dt[/tex]
[tex]F(x)=-((x^2+4x)-12)[/tex]
[tex]F(x)=12-4x-x^2[/tex]

Yet none of the results match (B is the closest, but it's still not correct).

Are you sure it's the integral [tex]\displaystyle\int_x^2[/tex], and not the other way around, [tex]\displaystyle\int_2^x[/tex]?

Answer:

Option B - [tex]F(x)=x^2+4x-12[/tex]

Step-by-step explanation:

Given : Use the graph of [tex]f(t)=2t+4[/tex] on the interval [-4, 6].

To find : Write the function F(x), where F(x)=the integral from x to 2 of f(t) dt?

Solution :

The function [tex]f(t)=2t+4[/tex]

To write F(x), we integrate the function from x to 2

i.e. [tex]F(x)=\int\limits^2_x {f(t)} \, dt[/tex]

[tex]F(x)=\int\limits^2_x {2t+4} \, dt[/tex]

[tex]F(x)=[2(\frac{t^2}{2})+4t]\limits^2_x[/tex]

[tex]F(x)=[t^2+4t]\limits^2_x[/tex]

[tex]F(x)=(2^2+4(2))-(x^2+4(x))[/tex]

[tex]F(x)=4+8-x^2-4x[/tex]

[tex]F(x)=12-x^2-4x[/tex]

[tex]F(x)=x^2+4x-12[/tex]

Therefore, Option B is correct.

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