Use a left and midpoint Rienman sum to approximate

[Mark44]

But the question said "right Reinman"

What's your point? All this means is that you use the right endpoint of each subinterval.

Pick a number n.
Divide the interval [0, 3] into that many subintervals.
Evaluate f at the [STRIKE]left end [/STRIKE] [STRIKE]midpoint[/STRIKE] right end of each subinterval.
Multiply by \Delta x.
Add the n terms.



BTW, it's Riemann.

 

[flyingpig]

What if you want to use left endpoint

 

[Sethric]

Try this:

Riemann Sum

Particularly the difference between right, midpoint, and left.

 

[Mark44]

What if you want to use left endpoint

See post #24.

I really hope that you have finished this problem by now. Approximating an integral by a Riemann sum is really nothing more than adding up the areas of a number of (n) rectangles. The rectangles are all \Delta x (= (b - a)/n) in width. For a left Riemann sum, you use the left endpoint of the subinterval to find the height of the rectangle. For a right Riemann sum, you use the right endpoint of the subinterval to find the height of the rectangle. For a midpoint Riemann sum, you use the midpoint of the subinterval to find the height of the rectangle.

To get your approximation, add the areas of the n rectangles. That's all there is to it.