# Thread: Continuous Function on an Interval (Cal 2)

1. ## Continuous Function on an Interval (Cal 2)

Let f(x) be a continuous function defined on the interval [2, infinity) such that
f(4) = 10, |f(x)| < x^{9} + 6
and

INT (f(x) e^(-x/2))dx = 6.
4

(integral from 4 to infinity of f(x) times e to the negative x over 2)

Determine the value of

INT (f '(x) e^(-x/2))dx
4

So, it gives a value for the definite integral, as well as an x,y coordinate of the function itself and an equation that f(x) is less than...and it asks for the same definite integral with the derivative of f(x) replacing f(x).

Not quite sure where to start (or end for that matter!), if anyone could point me in the right direction (quickly) I would appreciate it!

2. Originally Posted by habsfan
Let f(x) be a continuous function defined on the interval [2, infinity) such that
f(4) = 10, |f(x)| < x^{9} + 6
and

INT (f(x) e^(-x/2))dx = 6.
4

(integral from 4 to infinity of f(x) times e to the negative x over 2)

Determine the value of

INT (f '(x) e^(-x/2))dx
4

So, it gives a value for the definite integral, as well as an x,y coordinate of the function itself and an equation that f(x) is less than...and it asks for the same definite integral with the derivative of f(x) replacing f(x).

Not quite sure where to start (or end for that matter!), if anyone could point me in the right direction (quickly) I would appreciate it!
$\int f(x)e^{-x/2}dx$

if we integrate by parts with $u=f(x) \mbox{ then } du=f'(x)dx$
and $dv=e^{-x/2}dx \mbox{ then} v=-2e^{-x/2}$

$\int_{4}^{\infty} f(x)e^{-x/2}dx=-2e^{-x/2}f(x) |_{4}^{\infty}+2\int_{4}^{\infty} f'(x)e^{-x/2}dx$

use what is given to finish from here.

hint: $\lim_{x \to \infty} f(x)e^{-x/2}=0$ Why? think about it and good luck.