Integration

**slevvio** · December 12th 2009, 12:53 PM

Using the definition of integration $\int_a^b f(t)dt = sup\{\int_{- \infty}^{\infty} g(t)dt$ where g is a step function supported on [a,b], $g(x) \le f(x) \forall x \in [a,b] \}$ (f continuous on [a,b] )

show that for $\lambda \ge 0, \int_a^b \lambda f(t)dt = \lambda \int_a^b f(t)dt$ .

I would really appreciate any help with this. It's easy to show that if lambda is 0 you get the result but I am having trouble proving the rest. I just get tangled up in algebra, though I think I am trying to prove sup A = $\lambda$ sup B, where A = $\{\int_{- \infty}^{\infty} g(t)dt$ where g is a step function supported on [a,b], $g(x) \le \lambda f(x) \forall x \in [a,b] \}$ and B = $\{\int_{- \infty}^{\infty} g(t)dt$ where g is a step function supported on [a,b], $g(x) \le f(x) \forall x \in [a,b] \}$ Thank you very much.

**HallsofIvy** · December 13th 2009, 06:44 AM

$sup\{\lambda F(x)\}= \lambda sup\{ F(x)\}$

(and, of course, $\int \lambda F(x)dx= \lambda \int F(x)dx$ )

**Shanks** · December 13th 2009, 08:15 AM

provide $\lambda \geq 0$ :
for any step function g which supported on [a,b], $g(x) \le f(x) \forall x \in [a,b]$ , $\lambda g(x)$ is a step function supported on [a,b], and $\lambda g(x) \le \lambda f(x) \forall x \in [a,b]$ .
taking the superem over all step function g, we get that the LHS is no less than the RHS.
and similarly the RHS is no less than the LHS. QED

**slevvio** · December 13th 2009, 03:21 PM

thank you, worked it out

Math Help - Integration

LinkBack

Thread Tools

Display

Integration

Similar Math Help Forum Discussions

Integration help please these are three small even basic integration problems

sketch the region of integration and change the order of integration.

Definite Integration - Double integration by parts

Simple Integration problem: Trigonometric Integration

Tricky integration, possibly Integration Factor

Search Tags