Riemann sum

Question

 1. Evaluate the Riemann sum for (x) = x3 − 6x, for 0 ≤ x ≤ 3 with six sub intervals, taking the sample points, xi, to be the right endpoint of each interval. Give three decimal places in your answer.

Explain, using a graph of f(x), what the Riemann sum in Question #1 represents. 

Answer 1

First obtain for comparison the evaluation of ∫_{(x=0 to x=√6)}[x³ − 6x]dx which integrates

to {0.25x⁴ − 3x²|_{x=0 to x=√6}} equal to 0.25(√6)⁴ − 3(√6)² minus 0.25(0)⁴ + 3(0)² or 9 − 18 or -9.

Then add -9 to ∫_{(x=√6 to x=3)}[x³ − 6x]dx which integrates to {0.25x⁴ − 3x²|_{x=√6 to x=3}}

or 0.25(3)⁴ − 3(3)² − 0.25(√6)⁴ + 3(√6)² or 81/4 − 27 − 9 +18 or 2.25.

The net evaluation of ∫_{(x=0 to x=3)}[x³ − 6x]dx is then -9 + 2.25 or -6.75.

Demonstrate the Riemann Sum with a Programmable Calculator:

0→S---------------------------Register S starts at 0

-I→N---------------------------Register I is loaded with the desired increment; Register N is started at -I

Lbl 7---------------------------Mark the beginning of the "loop"

I + N→N----------------------Increase Register N by I

I(N³ − 6N) + S→S---------Multiply (N³ − 6N) by I and add the product to S

N>3⇒S.----------------------If N exceeds 3, show S and stop

Go to 7-----------------------For N ≤ 3, return to Lbl 7 and repeat the loop

A graph of x³ − 6x shows the curve dipping below the x-axis from x=0 to x=√6 and rising above the x-axis from x=√6 to x=3. The negative area is equal to -9 with the positive area at 2.25 for a net area of -6.75.

Running the program above with I = 0.001 gives a Riemann Sum of -6.736476741, very near to -6.75.

Running the program above with I = 0.01 gives a Riemann Sum of -6.61266599.

Running the program above with I = 0.1 gives a Riemann Sum of -5.1584.

Finally, running the program with I = 0.5 (6 equal intervals from x=0 to x=3) gives a Riemann Sum of 7.