Make-up #9

Riemann's Sums

Riemann's Sumvs help approximate areas under curves using rectangles or trapezoids. There are four different formulas each with different properties.

LRAM-left-hand approximation

▲x[f(a)+f(a+▲x)+...+f(b-▲x)]

http://upload.wikimedia.org/wikipedia/commons/c/c9/LeftRiemann2.svg

RRAM-right-hand approximation

▲x[f(a+▲x)+f(a+2▲x)+...+f(b)]

http://upload.wikimedia.org/wikipedia/commons/4/45/RightRiemann2.svg

MRAM-middle approximation

▲x[f(mid)+f(mid)+...+f(mid)]

http://upload.wikimedia.org/wikipedia/commons/d/d2/MidRiemann2.svg

Trapezoidal-trapezoid approximation (most accurate)

(▲x)/2[f(a)+2f(a+▲x)+2f(a+2▲x)+...+f(b)]

http://upload.wikimedia.org/wikipedia/commons/7/76/TrapRiemann2.svg

I bet your all wondering what ▲x is. For these types of problems your always given an interval [a,b].▲x is just (b-a)/n. N? yes n is the number of subintervals within your intervals. The question will usually tell you how many subintervals you have.

Say your giving a problem tell you to find the LRAM of the interval [0,4] with four subintervals. Since your ▲x=(b-a)/n you would subtract 0 from 4 and then divide that number by 4 giving you ▲x=1.

So now that you know that, you can solve for LRAM.

▲x[f(a)+f(a+▲x)+f(a+2▲x)+f(a+3▲x)]=LRAM.

1[(0)+(1)+(2)+(3)]=6