Question

Express the integral as a limit of Riemann sums... (image below)

Answer 1

That question was answered!

Okay, on to the next one. "Use the comparison properties of integrals to verify the inequality without evaluating the integrals."\[\int_{0}^{\pi/2}(\sin{x}+\cos{x})dx\ge0\]

Answer 2

@Zepdrix

Answer 3

What's a comparison property

Answer 4

I dunno =O ill have to google it

Answer 5

hmm

Answer 6

Ok ok ok I think I see what's going on.

Answer 7

If our function can be given bounds at both ends of the interval values,
\(\large\rm m\le f(x)\le M\)

then,
\(\large\rm m\left(b-a\right)\le\int\limits_a^b f(x)dx\le M\left(b-a\right)\)
the integral is bounded by these rectangles.

Answer 8

sinx+cosx at 0 is ... 1, right?
How bout the other boundary?

Answer 9

Oh.. Hmm it's 1 at both ends.

Answer 10

Eh

Answer 11

I really don't know. What is this comparison thing anyway?

Answer 12

Ok it actually has a max value at pi/4.
sin(pi/4)+cos(pi/4) = sqrt2

Answer 13

Ok

Answer 14

So we can say,
\(\large\rm 1\le sin x+cos x\le \sqrt2\)

Answer 15

On the given interval.

Answer 16

So then, the integral will be bounded by these rectangles,

\(\large\rm \large\rm 1\left(\frac{\pi}{2}-0\right)\le \int\limits_0^{\pi/2}sin x+cos x\le \sqrt2\left(\frac{\pi}{2}-0\right)\)

Answer 17

So we've shown that our integral is between two positive values.

Answer 18

I think that's how we wanted to do that...
This property is weird. I've never seen it before :d

Answer 19

Sorry, working on multiple problems. I will be back soon ...>_>