Question

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

Answer 1

When
`x<2`, you have


`F(x)=\displaystyle\int_x^2(2t+4)\,\mathrm dt`

`F(x)=(t^2+4t)\bigg|_(t=x)^(t=2)`

`F(x)=(2^2+4(2))-(x^2+4x)`

`F(x)=12-4x-x^2`

Even when
`x>2`, you would get the same answer.


`F(x)=\displaystyle\int_x^2(2t+4)\,\mathrm dt=-\int_2^x(2t+4)\,\mathrm dt`

`F(x)=-((x^2+4x)-12)`

`F(x)=12-4x-x^2`

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

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

Answer 2

Option B -
`F(x)=x^2+4x-12`

Explanation:

Given : Use the graph of
`f(t)=2t+4` 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
`f(t)=2t+4`

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

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

`F(x)=\int\limits^2_x {2t+4} \, dt`

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

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

`F(x)=(2^2+4(2))-(x^2+4(x))`

`F(x)=4+8-x^2-4x`

`F(x)=12-x^2-4x`

`F(x)=x^2+4x-12`

Therefore, Option B is correct.