A definite integral is not just an area but a signed area.
Area above the x-axis counts as positive and area below as negative, so when both are present they can cancel.
The key to computing it is the fundamental theorem of calculus: find an antiderivative F and subtract its value at the lower limit, F(a), from its value at the upper limit, F(b).
In the first picture, moving a and b shows how the positive and negative parts of sin x add and cancel; in the second, the area under 1/x² never passes 1, however far the upper limit goes.
Definite integrals and signed area
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y = sin x from a to b: the part above the x-axis (gold) is added and the part below (blue) is subtracted
0
3.1
👀 Signed area
①Area above the x-axis counts as +, area below as −
②With a = 0 and b = π the integral is 2 (all above the axis)
③With a = 0 and b = 2π it is 0 (above and below cancel)
④b moves in steps of 0.1 in the picture, so the value comes close to 2 at 3.1 rather than at π
Fundamental Theorem of Calculus
Fundamental theorem of calculus
∫ab f(x) dx = F(b) − F(a) (F' = f)
Definite integral = antiderivative at the upper limit − antiderivative at the lower limit
🌉 The bridge between derivatives and integrals
①Differentiation and integration undo each other
②Find F with F'(x) = f(x), and the definite integral is F(b) − F(a)
③For sin x in the picture, F(x) = −cos x, so the integral is cos a − cos b
The theorem in another form
ddx ∫ax f(t) dt = f(x)
Differentiating an integral whose upper limit is x gives back the integrand
Properties of Definite Integrals
Splitting the interval
∫ab f dx = ∫ac f dx + ∫cb f dx
Integrating over pieces and adding gives the integral over the whole interval
Functions symmetric about the y-axis
f(−x) = f(x) → ∫−aa f dx = 2∫0a f dx
If the graph is symmetric about the y-axis, compute half and double it
Functions symmetric about the origin
f(−x) = −f(x) → ∫−aa f dx = 0
If the graph is symmetric about the origin, the integral over a symmetric interval is 0
⚖️ Symmetry saves work
①For a function symmetric about the y-axis (f(−x) = f(x)), integrate from 0 to a and double
②For a function symmetric about the origin (f(−x) = −f(x)), the two sides cancel to 0
③For a polynomial, split it into even-power and odd-power terms to use these facts
Integrals to infinity (advanced)
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The area under y = 1/x² from 1 to t; however large t gets, the area 1 − 1/t only approaches 1 and never passes it
5
∞ Endless, yet finite area
①1/x² shrinks so fast that its area stays finite even though it goes on forever
②The area from 1 to t is 1 − 1/t, which approaches 1 as t → ∞
③By contrast, the area under 1/x from 1 to t is ln t, which grows without bound as t → ∞
When the area is finite
∫1∞1xp dx : p > 1 converges, p ≤ 1 diverges
For p > 1 the area is finite; for p ≤ 1 it grows without bound. This goes beyond the usual high-school course
Worked examples
Example 1
Evaluate the definite integral ∫12 (3x2 + 2) dx.
1
Take F(x) = x³ + 2x, a function whose derivative is 3x² + 2.
F(x) = x3 + 2x
2
By the fundamental theorem, compute F(2) − F(1).
(23 + 2·2) − (13 + 2·1) = 12 − 3 = 9
▸ 9
F(2) = 12 and F(1) = 3, so the answer is 9. The constant C of the antiderivative cancels in the subtraction.
Example 2
Evaluate ∫−22 (x3 + x2) dx.
1
x³ is symmetric about the origin (integral 0) and x² about the y-axis (compute half and double), so split the two.
∫−22 x3 dx = 0, ∫−22 x2 dx = 2∫02 x2 dx
2
Compute only the x² part, which is symmetric about the y-axis.
2 · 233 = 163
▸ 163
On a symmetric interval the part symmetric about the origin vanishes and the part symmetric about the y-axis needs only half the work, which saves a lot of calculation.
Summary
FTC
∫ab f(x) dx = F(b) − F(a), F'(x) = f(x)
Computing a definite integral comes down to finding an antiderivative
CSAT-style problem, adapted
For g(x) = ∫1x (2t + 1) dt, what is g'(2) + g(1)?
①3
②4
③5
④6
⑤7
▸ ③ 5
1
By the fundamental theorem, g'(x) = 2x + 1, so g'(2) = 5.
g'(x) = 2x + 1 ⇒ g'(2) = 5
2
g(1) is the integral from 1 to 1, which is 0.
g(1) = 0 ⇒ g'(2) + g(1) = 5
🎯 Exam Points
①Computing F(b) − F(a) is the foundation
②Parts below the x-axis count as negative, so do not confuse the integral with area
③On symmetric intervals, use symmetry to cut the work
④Differentiating an integral with upper limit x gives the integrand
⑤An integral with upper limit x is 0 when x equals the lower limit