Calculate `int_1^2 f(x)dx `  if `f(x)=2xF(x), f(1)=2e` domain=range `(0,oo)`

1 Answer | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You should remember the equation that relates f(x) and F(x) such that:

`F'(x) = f(x)`

The problem provides the information that  `f(x)=2xF(x), ` hence, you should substitute F'(x) for f(x) such that:

`F'(x) = 2xF(x) =gt 2x = (F'(x))/(F(x))`

Integrating both sides yields:

`int 2x dx= int (F'(x))/(F(x)) dx`

`2x^2/2 = ln F(x) =gt x^2 + c= ln F(x) =gt F(x) = e^(x^2 + c)`

Notice that the problem provides the information that f(1)=2e, hence, you should substitute 1 for x in `f(x)=2xF(x),`   such that:

`f(1) = 2F(1) =gt f(1) = 2e^(1 + c)`

But `f(1)=2e =gt 2e^(1 + c) = 2e =gt 1 + c = 1 =gt c = 0`

Hence, `F(x) = e^(x^2)`

Since you know F(x), you may evaluate the definite integral such that:

`int_1^2 f(x) dx= F(2) - F(1)`

`int_1^2 f(x) dx = e^(2^2) - e`

Factoring out `e`  yields:

`int_1^2 f(x) dx = e(e^3 - 1)`

Hence, evaluating the definite integral under the given condition yields `int_1^2 f(x) dx = e(e^3 - 1).`

We’ve answered 317,805 questions. We can answer yours, too.

Ask a question