- #1

LadiesMan

- 96

- 0

Attempt:

Since 2 f(x) is greater than f(x) we can call it g(x) and that will be the dominant function.

[tex]\int (g(x) - f(x)) dx[/tex]

It becomes...

[tex]G(x) - F(x)[/tex]

What do I do next?

Thanks

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- Thread starter LadiesMan
- Start date

- #1

LadiesMan

- 96

- 0

Attempt:

Since 2 f(x) is greater than f(x) we can call it g(x) and that will be the dominant function.

[tex]\int (g(x) - f(x)) dx[/tex]

It becomes...

[tex]G(x) - F(x)[/tex]

What do I do next?

Thanks

- #2

Pere Callahan

- 586

- 1

Note that you want to consider definite integrals

[tex]

\int_a^b{dx(g(x)-f(x))}

[/tex]

Can you simplify g(x)-f(x)..?

- #3

LadiesMan

- 96

- 0

umm yeah i guess that took me off course.

- #4

LadiesMan

- 96

- 0

so then how would i do it?

- #5

Pere Callahan

- 586

- 1

Uhm,

g(x)-f(x) = 2f(x)-f(x) =...?

g(x)-f(x) = 2f(x)-f(x) =...?

- #6

LadiesMan

- 96

- 0

Umm that equals f(x)

- #7

Pere Callahan

- 586

- 1

very good, so

[tex]\int_a^b{dx(g(x)-f(x))}=\int_a^b{dx(2f(x)-f(x))}=\int_a^b{dx(f(x))}=...=?[/tex]

[tex]\int_a^b{dx(g(x)-f(x))}=\int_a^b{dx(2f(x)-f(x))}=\int_a^b{dx(f(x))}=...=?[/tex]

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