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high school Definite Integral

The area under a curve has no ready-made formula like the area of a rectangle or a triangle. So the region is filled with thin rectangles to estimate the area, and the rectangles are made thinner and thinner. As they get thinner the estimate approaches the true area, and that limit is the definite integral. If an antiderivative F is known, the limit can be computed as simply F(b) − F(a). In the first picture, raising the number of strips n brings the sum of the rectangles toward the exact value b³/3; the second shows that the parts below the x-axis are subtracted.

Cutting the area into thin rectangles

📐 The idea behind the definite integral
①Estimate the area under a curve by lining up many thin rectangles
②Making them thinner and thinner (n → ∞) brings the estimate toward the true area
③That limit is the definite integral

From sums of rectangles to the definite integral

5
2
Definition
∫ab f(x) dx = limn→∞ Σk=1n f(xk) Δx
The limit of the sum of rectangles of width Δx = (b − a)/n is the area under the curve
🔑 What to look for
①With small n the rectangles are wide and miss the curve by a lot
②From about n = 20 the rectangles fill the region almost exactly
③As n grows without bound the sum approaches the exact value b³/3
④Using the height at the middle of each strip lets overshoot and shortfall cancel, so the error shrinks quickly

Fundamental Theorem of Calculus

FTC
∫ab f(x) dx = F(b) - F(a) (with F'(x) = f(x))
Knowing an antiderivative gives the definite integral at once, with no sums of rectangles
💡 What the fundamental theorem does
①Computing the limit of sums directly is very laborious
②By the fundamental theorem, find F with F'(x) = f(x) and compute F(b) − F(a)
③It means differentiation, which deals with slope, and integration, which deals with area, undo each other
④The integral from a to c plus the integral from c to b is the integral from a to b
⑤If the graph is symmetric about the y-axis, the integral from −a to a is twice the one from 0 to a; if it is symmetric about the origin, it is 0
Worked Example
∫02 x² dx = [x³3]02 = 83 - 0 = 83
Find F(x) = x³/3, then compute F(2) − F(0); it matches the exact value in the first picture when b = 2

Definite integrals and the sign of area

1.5
Area and the definite integral
Area S = ∫ab |f(x)| dx (absolute value!)
A definite integral is a signed value; the actual area is the integral of |f(x)|
⚠ Points to watch
①The part below the x-axis counts as negative, so the integral can differ from the area
②With b = 2 in the picture, the areas above and below are equal, so the integral is 0 even though the area is not
③For the actual area, split the interval where the graph meets the x-axis and add the absolute values of the parts

Summary

Core
∫ab f(x) dx = F(b) - F(a)
Fundamental Theorem of Calculus
🎯 Exam Points
①A definite integral is a signed area and can differ from the area
②Fundamental theorem: the difference of antiderivatives F(b) − F(a)
③Splitting the interval and adding gives the same value
④For a function symmetric about the y-axis, the integral from −a to a is twice the one from 0 to a
⑤For a function symmetric about the origin, the integral from −a to a is 0

Worked Examples & Past Exam

Example 1
Evaluate ∫02 3x² dx.
1
Take F(x) = x³, a function whose derivative is 3x².
∫ 3x2 dx = x3
2
By the fundamental theorem, compute F(2) − F(0).
[x3]02 = 23 - 03 = 8
▸ 8
The constant C of an antiderivative cancels in F(b) − F(a), so it can be left out when computing a definite integral.
Example 2
Evaluate ∫13 (2x − 1) dx.
1
Take F(x) = x² − x, a function whose derivative is 2x − 1.
∫ (2x - 1) dx = x2 - x
2
Compute F(3) − F(1).
[x2 - x]13 = (9 - 3) - (1 - 1) = 6
▸ 6
Under y = 2x − 1 lies a trapezoid of height 1 at x = 1 and 5 at x = 3, so (1 + 5) × 2 ÷ 2 = 6 confirms it.
2022 KICE mock exam Math type, adapted
What is ∫01 (3x² + 2) dx?
①3
②1
③5
④2
⑤4
▸ ① 3
1
Take F(x) = x³ + 2x, a function whose derivative is 3x² + 2.
∫ (3x2 + 2) dx = x3 + 2x
2
Compute F(1) − F(0).
[x3 + 2x]01 = (1 + 2) - 0 = 3
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Applications of Definite Integrals

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