Riemann Sums Formula

Riemann sums are a method of approximating the definite integral _a^b f(x)\,dx by dividing the interval [a, b] into subintervals and summing the areas of.

The Formula

∫abf(x) dxβ‰ˆβˆ‘i=1nf(xiβˆ—) ΔxwhereΒ Ξ”x=bβˆ’an\int_a^b f(x)\,dx \approx \sum_{i=1}^{n} f(x_i^*)\,\Delta x \quad \text{where } \Delta x = \frac{b-a}{n}
xiβˆ—x_i^* is the sample point (left, right, or midpoint) in each subinterval.

When to use: Imagine filling the area under a curve with thin rectangles. The more rectangles you use, the better the approximation. In the limit of infinitely many infinitely thin rectangles, you get the exact areaβ€”which is the definite integral.

Quick Example

Approximate ∫02x2 dx\int_0^2 x^2\,dx using 4 right-endpoint rectangles:
Ξ”x=0.5,βˆ‘=0.5[0.25+1+2.25+4]=0.5Γ—7.5=3.75\Delta x = 0.5, \quad \sum = 0.5[0.25 + 1 + 2.25 + 4] = 0.5 \times 7.5 = 3.75
Exact value: 83β‰ˆ2.667\frac{8}{3} \approx 2.667. The overestimate is because x2x^2 is increasing.

Notation

LnL_n = left Riemann sum, RnR_n = right Riemann sum, MnM_n = midpoint sum, TnT_n = trapezoidal sum.

What This Formula Means

A method of approximating the definite integral ∫abf(x) dx\int_a^b f(x)\,dx by dividing the interval [a,b][a, b] into subintervals and summing the areas of rectangles (or trapezoids) whose heights are determined by the function.

Imagine filling the area under a curve with thin rectangles. The more rectangles you use, the better the approximation. In the limit of infinitely many infinitely thin rectangles, you get the exact areaβ€”which is the definite integral.

Formal View

Let P={x0,x1,…,xn}P = \{x_0, x_1, \ldots, x_n\} be a partition of [a,b][a,b] with Ξ”xi=xiβˆ’xiβˆ’1\Delta x_i = x_i - x_{i-1} and xiβˆ—βˆˆ[xiβˆ’1,xi]x_i^* \in [x_{i-1}, x_i]. The Riemann sum is S(P,f)=βˆ‘i=1nf(xiβˆ—)Ξ”xiS(P, f) = \sum_{i=1}^{n} f(x_i^*) \Delta x_i. Then ∫abf(x) dx=lim⁑βˆ₯Pβˆ₯β†’0S(P,f)\int_a^b f(x)\,dx = \lim_{\|P\| \to 0} S(P, f) where βˆ₯Pβˆ₯=max⁑iΞ”xi\|P\| = \max_i \Delta x_i.

Worked Examples

Example 1

easy
Approximate ∫02x2 dx\int_0^2 x^2\,dx using a left Riemann sum with n=4n = 4 equal subintervals.

Answer

L4=1.75L_4 = 1.75 (underestimate; exact =83= \frac{8}{3})

First step

1
Ξ”x=0.5\Delta x = 0.5. Left endpoints: x=0,0.5,1,1.5x = 0, 0.5, 1, 1.5.

Full solution

  1. 2
    f(0)=0,β€…β€Šf(0.5)=0.25,β€…β€Šf(1)=1,β€…β€Šf(1.5)=2.25f(0)=0,\; f(0.5)=0.25,\; f(1)=1,\; f(1.5)=2.25.
  2. 3
    L4=0.5(0+0.25+1+2.25)=0.5Γ—3.5=1.75L_4 = 0.5(0+0.25+1+2.25) = 0.5 \times 3.5 = 1.75.
  3. 4
    Exact value: 83β‰ˆ2.667\frac{8}{3} \approx 2.667. The left sum underestimates because ff is increasing.
For an increasing function, left endpoints give the minimum in each subinterval, so the left Riemann sum underestimates. More subintervals improve accuracy.

Example 2

medium
Approximate ∫131x dx\int_1^3 \frac{1}{x}\,dx using a right Riemann sum with n=4n = 4 subintervals and classify the estimate.

Example 3

medium
Approximate ∫04x2 dx\int_0^4 x^2\,dx using a midpoint sum with n=4n=4.

Common Mistakes

  • Forgetting Ξ”x=bβˆ’an\Delta x=\frac{b-a}{n} β€” the rectangle width must come from dividing the interval, not assumed to be 11.
  • Using the wrong sample point β€” left, right, and midpoint heights differ; match the height to the rule requested.
  • Calling a finite sum the exact area β€” it's an approximation until you take the limit as nβ†’βˆžn\to\infty.

Why This Formula Matters

Riemann sums are the definition of the definite integral: the exact integral is the limit as rectangles become infinitely thin. They're also the practical tool when a function has no elementary antiderivative (like ex2e^{x^2}), so you can't use FTC and must estimate numerically. Recognizing it by "Am I summing rectangle areas f(xiβˆ—)Ξ”xf(x_i^*)\Delta x to estimate area under a curve, rather than evaluating exactly?" β€” rather than by familiar numbers β€” is what lets a student tell it apart from definite integral (exact) and trapezoidal rule and left vs right vs midpoint in a mixed problem set.

Frequently Asked Questions

What is the Riemann Sums formula?

A method of approximating the definite integral ∫abf(x) dx\int_a^b f(x)\,dx by dividing the interval [a,b][a, b] into subintervals and summing the areas of rectangles (or trapezoids) whose heights are determined by the function.

How do you use the Riemann Sums formula?

Imagine filling the area under a curve with thin rectangles. The more rectangles you use, the better the approximation. In the limit of infinitely many infinitely thin rectangles, you get the exact areaβ€”which is the definite integral.

What do the symbols mean in the Riemann Sums formula?

LnL_n = left Riemann sum, RnR_n = right Riemann sum, MnM_n = midpoint sum, TnT_n = trapezoidal sum.

Why is the Riemann Sums formula important in Math?

Riemann sums are the definition of the definite integral: the exact integral is the limit as rectangles become infinitely thin. They're also the practical tool when a function has no elementary antiderivative (like ex2e^{x^2}), so you can't use FTC and must estimate numerically. Recognizing it by "Am I summing rectangle areas f(xiβˆ—)Ξ”xf(x_i^*)\Delta x to estimate area under a curve, rather than evaluating exactly?" β€” rather than by familiar numbers β€” is what lets a student tell it apart from definite integral (exact) and trapezoidal rule and left vs right vs midpoint in a mixed problem set.

What do students get wrong about Riemann Sums?

The procedure for riemann sums is the easy part; the trap is forgetting Ξ”x=bβˆ’an\Delta x=\frac{b-a}{n}. Asking "Am I summing rectangle areas f(xiβˆ—)Ξ”xf(x_i^*)\Delta x to estimate area under a curve, rather than evaluating exactly?" first is what keeps a correct-looking calculation from being attached to the wrong concept.

What should I learn before the Riemann Sums formula?

Before studying the Riemann Sums formula, you should understand: integral, definite integral.

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How to Integrate Rational Functions: Long Division and Partial Fractions β†’
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Related Concepts

IntegralDefinite Integral