Riemann sums to the Fundamental Theorem of Calculus: why the area under a curve is an antiderivative, with one integral done both ways

A definite integral is the limit of Riemann sums: slice the region under a curve into n rectangles, add their areas, and let n grow. The Fundamental Theorem of Calculus says you never have to take that limit: the exact area from a to b is F(b) − F(a) for any antiderivative F of the function. This page computes one integral both ways, the sums with four rectangles and then the theorem, so you see the sums closing in on the exact value, and names the one line students get wrong when they evaluate.

Riemann sums: the area in rectangles

To estimate the area under y = f(x) from x = a to x = b, cut the interval into n slices of equal width. Each slice has width Δx = (b − a)/n. On each slice, stand a rectangle whose height is the function's value at one chosen point of the slice:

• at the left end, for a left Riemann sum,
• at the right end, for a right Riemann sum,
• at the middle, for the midpoint rule.

Add the rectangles' areas. Written once in sigma notation, the sum is Σ f(xᵢ) Δx, where xᵢ is the chosen point of slice i. In words: add up height times width, slice by slice.

The three choices give different answers. For a function that increases across the interval, the left end of each slice is its lowest point, so the left sum sits under the curve and underestimates. The right end is the highest point, so the right sum overestimates. The midpoint usually lands close.

In a Calculus I course this is idea 20, Riemann sums and the definite integral, and the theorem is idea 21. MCV4U has no integration, so an Ontario student meets both in first year: first-year calculus says what changes. Both are also central to AP Calculus.

Worked example, part 1: three Riemann sums for ∫₀² x² dx with n = 4

The function is f(x) = x², the interval is from 0 to 2, and there are 4 slices. So Δx = (2 − 0)/4 = 0.5, and the slice boundaries are 0, 0.5, 1, 1.5 and 2.

Right Riemann sum. The right ends are 0.5, 1, 1.5 and 2.

• Heights: 0.5² = 0.25, 1² = 1, 1.5² = 2.25, 2² = 4.
• Add the heights: 0.25 + 1 + 2.25 + 4 = 7.5.
• Multiply by the width: 7.5 × 0.5 = 3.75.

Left Riemann sum. The left ends are 0, 0.5, 1 and 1.5.

• Heights: 0² = 0, 0.5² = 0.25, 1² = 1, 1.5² = 2.25.
• Add the heights: 0 + 0.25 + 1 + 2.25 = 3.5.
• Multiply by the width: 3.5 × 0.5 = 1.75.

Midpoint rule. The midpoints are 0.25, 0.75, 1.25 and 1.75.

• Heights: 0.25² = 0.0625, 0.75² = 0.5625, 1.25² = 1.5625, 1.75² = 3.0625.
• Add the heights: 0.0625 + 0.5625 + 1.5625 + 3.0625 = 5.25.
• Multiply by the width: 5.25 × 0.5 = 2.625.

The common wrong lines

Reporting 7.5 as the right sum. That is the total of the heights, not the area. An area is height times width, so every sum ends with "× Δx".

Using five heights. Four slices have five boundary points, and it is easy to evaluate the function at all five. But each rectangle takes exactly one height, so four slices take four heights: drop the first point for a right sum, the last for a left sum.

Worked example, part 2: the Fundamental Theorem on the same integral

Now the exact value, by the Fundamental Theorem of Calculus.

• Line 1, find an antiderivative. An antiderivative of x² is F(x) = x³/3, because the derivative of x³/3 is x².
• Line 2, evaluate at the upper limit, then the lower, and subtract. F(2) − F(0) = 8/3 − 0 = 8/3 ≈ 2.667.
• Line 3, compare with the sums.

Method (n = 4)	Value
Left Riemann sum	1.75
Midpoint rule	2.625
Exact, F(2) − F(0)	8/3 ≈ 2.667
Right Riemann sum	3.75

The exact value sits between the left and right sums, as it must for an increasing function, and the midpoint sum is already within about 0.04 of it with only four slices. As n grows, all three sums close in on 8/3. That limit is what "definite integral" means, and the theorem hands it to you in one subtraction.

There is a fault this example hides. Here F(0) = 0, so a student who writes F(2) alone still gets 8/3. The next example has a lower limit that is not zero, which is the only way to see whether the subtraction is really happening.

How to evaluate a definite integral, line by line

Evaluate ∫₁³ (x² + 1) dx.

• Line 1, find an antiderivative. F(x) = x³/3 + x.
• Line 2, evaluate at the upper limit. F(3) = 27/3 + 3 = 9 + 3 = 12.
• Line 3, evaluate at the lower limit. F(1) = 1/3 + 1 = 4/3.
• Line 4, subtract, upper minus lower. F(3) − F(1) = 12 − 4/3 = 36/3 − 4/3 = 32/3 ≈ 10.667.

The common wrong lines

Adding + C. If you write F(x) = x³/3 + x + C, the C appears in F(3) and again in F(1) and cancels in the subtraction. It does no arithmetic harm, but writing it signals that the definite integral (a number) and the indefinite integral (a family of functions) are being confused.

Reporting F(3) = 12 alone. This is the fault the first example hid. The lower limit has a value, and it must be subtracted.

Subtracting in the wrong order: F(1) − F(3) = −32/3. The integrand x² + 1 is positive and the interval runs forward from 1 to 3, so the area is positive. A negative answer here is a reversed subtraction.

Siblings to try cold

Cover the examples above and work each of these on paper first.

• The right Riemann sum for ∫₀³ x² dx with n = 3. Then compare it with the exact value.
• ∫₁² (3x² − 2) dx.

Check

• Δx = 1, the right ends are 1, 2 and 3, the heights are 1, 4 and 9, and the sum is (1 + 4 + 9) × 1 = 14. The exact value is 9, so the right sum overestimates, as it should for an increasing function.
• An antiderivative is x³ − 2x. At 2 it is 8 − 4 = 4; at 1 it is 1 − 2 = −1. So the integral is 4 − (−1) = 5. If you got 3, you subtracted 1 instead of −1: the lower value was negative, and subtracting a negative adds.

Why the area is the antiderivative: the Fundamental Theorem of Calculus, part 1 and part 2, in words

The theorem has two parts, and the second follows from the first.

Part 1. Let f be continuous. Define the area function A(x) as the area under f from a fixed point a up to x: A(x) = ∫ₐˣ f(t) dt. Its derivative is f(x). In words: move x a tiny distance h to the right, and you add a thin slice of width h and height about f(x), so the area grows by about f(x) · h. Divide by h and let h shrink, and the rate of change of the area is the height of the curve. This even works when no formula for the antiderivative exists: the derivative of ∫₀ˣ sin(t²) dt is simply sin(x²).

Part 2. Part 1 says the area function is an antiderivative of f. Any other antiderivative F differs from it by a constant, which is a consequence of the Mean Value Theorem and the + C. So F(b) − F(a) equals A(b) − A(a), and the constant cancels. A(a) is the area from a to a, which is 0, so A(b) − A(a) is the area from a to b. Where the curve dips below the x-axis, that part counts as negative, so strictly the theorem gives the signed area; for a curve above the axis, as in every example here, it is the area.

That is why the limit of the Riemann sums and F(b) − F(a) are the same number, and the table above showed it numerically. The full proof is in the textbook chapter; what you need for the course is this argument in words and the evaluation habit.

Where this sits in the course, and where it goes next

In Calculus I, idea 20, Riemann sums and the definite integral, and idea 21, The Fundamental Theorem of Calculus, come after idea 19, Antiderivatives and the indefinite integral, and before idea 22, Integration by substitution. The sum's "let n grow" is a limit, so if limits are shaky, the definition will be too.

The theorem turns every area into an antiderivative, which makes finding antiderivatives the next job: substitution and integration by parts are the two main techniques. After them, idea 24, Area between curves, is this same subtraction with two functions. Error bounds for the midpoint and trapezoidal rules also come later and are not needed here.

To make the evaluation habit automatic, test yourself rather than rereading this page: cover the worked examples and redo them cold. Roediger and Karpicke (2006) found that students who took practice tests remembered more a week later than students who restudied, and test yourself before you reread has the method. The example-then-sibling order follows Sweller and Cooper (1985), explained in worked examples, then practice problems.

Riemann sums and the Fundamental Theorem on one screen

• Riemann sum: Δx = (b − a)/n; add height × width for each slice, Σ f(xᵢ) Δx.
• Three choices of height: left end, right end, midpoint. For an increasing function, left underestimates and right overestimates.
• ∫₀² x² dx, n = 4: left 1.75, midpoint 2.625, exact 8/3 ≈ 2.667, right 3.75.
• Part 1: the area function's derivative is the function.
• Part 2: the exact area from a to b is F(b) − F(a), for any antiderivative F.
• Three evaluation faults: + C on a definite integral, F(b) alone, lower minus upper.