Key takeaway: A Riemann sum approximates the area under a curve by splitting the interval into subintervals, building a rectangle on each, and adding the rectangle areas.
Grade 12 – AP Calculus ABAP Calculus AB Unit 6Calculus
Definition of Riemann Sum
A Riemann sum estimates an area that has no easy formula. Divide the interval into pieces of width \(\Delta x\), choose a sample input in each piece to set a rectangle’s height, and add up height times width for every piece.
The choice of sample point names the rule: the left end, the right end, or the midpoint. More rectangles give a better estimate, and the limit as their number grows without bound is the definite integral.
Formula and notation
For a function \(r\) on \([m,n]\) with \(N\) equal pieces, each of width \(\Delta x=\frac{n-m}{N}\):
\[ \sum_{i=1}^{N} r(x_i^{*})\,\Delta x, \]
where \(x_i^{*}\) is the sample point in the \(i\)-th piece. For an increasing function, left sums underestimate and right sums overestimate.
Key facts
- The width of each piece is the interval length divided by the number of pieces.
- For an increasing function the left sum is too small and the right sum is too large.
- A midpoint sum is usually the most accurate of the three simple rules.
- Heights below the axis contribute negative amounts to the sum.
Riemann Sum example
Estimate the area under \(r(x)=0.5x^2+2\) on \([0,8]\) with a left Riemann sum using 4 equal pieces.
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Width: \(\Delta x=\frac{8-0}{4}=2\). Left endpoints: 0, 2, 4, 6.
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Heights: \(r(0)=2\), \(r(2)=4\), \(r(4)=10\), \(r(6)=20\).
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Sum: \(2\,(2+4+10+20)=2\cdot 36=72\).
The left Riemann sum is 72, an underestimate; the exact area is about 101.3.
Common mistake
Watch out
A student adds the four heights, \(2+4+10+20=36\), and reports 36 as the area estimate.
Why it is wrong
Each term of a Riemann sum is height times width, the area of one rectangle. Leaving out the width \(\Delta x=2\) gives only a sum of heights. Multiply the total height 36 by 2 to reach the estimate of 72.
Related terms and lessons
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