What the limits do.
∫ 2x dx asks for every function whose derivative is 2x — a family, needing + C. ∫₀³ 2x dx asks something sharper: the area trapped between the curve and the x-axis from x = 0 to x = 3. Limits turn the family into a single number.
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Area Under a Curve: The Definite Integral
Put limits on an integral and it stops being a family of functions — it becomes one number: the area between the curve and the x-axis.
The problem
∫₀³ 2x dx
The antiderivative of 2x is x²; evaluate it at the limits: 3² − 0² = 9. And geometry agrees — the shaded region is a triangle with base 3 and height 6: ½ × 3 × 6 = 9. B multiplies base by height without halving.
Evaluate: antiderivative at the top, minus at the bottom.
∫₀³ 2x dx = [x²]₀³
antiderivative first — no C needed
= 3² − 0²
top limit in, bottom limit out
= 9
one number — the area
Why no + C? It would appear in both evaluations and cancel: (9 + C) − (0 + C) = 9. Definite integrals shed the constant automatically.
A root is a power too. ∫ from 4 to 9 of √x dx: √x = x1/2, whose antiderivative is (23)x3/2. Top minus bottom: (23)(27) − (23)(8) = (23)(19) = 383, since 93/2 = 3³ = 27 and 43/2 = 2³ = 8.
Check it with plain geometry.
the region is a triangle
base 3 along the axis, height 6 at x = 3
A = ½ × 3 × 6 = 9 ✓
calculus and geometry agree
This is the whole point of the definite integral: for shapes geometry already knows, it reproduces the same answer — and then it keeps working for curved regions no area formula can touch.
Below the axis counts negative.
| The curve is… | The integral counts it… |
|---|---|
| above the x-axis | positive |
| below the x-axis | negative |
∫₋₃⁰ 2x dx = 0 − 9 = −9: same triangle, below the axis, so it counts negative. Asked for the total shaded area of a region that dips under the axis, split the integral at the crossing and flip the sign of the underwater part.
Variations: a negative lower limit, an unknown limit, limits upside down.
A negative lower limit. Evaluate ∫ from −1 to 2 of (3x² − 2) dx.
F(x) = x³ − 2x
antiderivative
F(2) = 8 − 4 = 4
top
F(−1) = (−1)³ − 2(−1) = −1 + 2 = 1
bottom, in brackets
4 − 1 = 3
top minus bottom
What is different: the bottom value has signs inside it, and then a minus in front of it as well.
F(−1) is not −3
Writing −1 − 2 = −3 forgets that −2 × (−1) is +2. That slip turns the answer into 4 − (−3) = 7.
An unknown upper limit. Find k > 1 if ∫ from 1 to k of 2x dx = 15.
[x²] from 1 to k = k² − 1
the bottom limit is not 0 this time
k² − 1 = 15 → k² = 16
solve
k = 4
−4 is not greater than 1
What is different: the limit is the unknown. Forgetting the “− 1” from the bottom limit gives k = √15.
Limits upside down. Evaluate ∫ from 3 to 1 of 2x dx.
[x²] from 3 to 1 = 1² − 3²
still top-written minus bottom-written
= 1 − 9 = −8
negative, though 2x > 0 there
What is different: swapping the limits flips the sign. The region (area 8) is above the axis; only the direction of travel changed.
Find the crossings first: missing limits, and total area.
(a) Find the area bounded by y = 4x − x² and the x-axis. (b) Find the total area between y = x² − 1 and the x-axis from x = 0 to x = 2.
(a) 4x − x² = x(4 − x) = 0 → x = 0, 4
“bounded by the axis” — the limits are the x-intercepts
[2x² − x³3]₀⁴ = 32 − 643 = 323
above the axis the whole way
(b) x² − 1 = 0 at x = 1
below the axis on 0 to 1, above on 1 to 2
∫₀¹ = (13 − 1) − 0 = −23 → area 23
antiderivative x³3 − x; flip the sign
∫₁² = (83 − 2) − (13 − 1) = 43
already positive
total = 23 + 43 = 2
add the sizes
One integral straight across gives the net, not the area
∫₀² (x² − 1) dx = 43 − 23 = 23 lets the underwater part cancel some of the rest. “Area” or “shaded region” always means sizes added; only “evaluate the integral” wants the signed value.
On the exam
Expect a definite integral to evaluate, a shaded area from a graph, or an unknown upper limit k. Sketch or factor to find where the curve meets the axis, then split there. Fast check: compare with a rectangle or triangle drawn around the region — the answer must be smaller than the box that contains it.
Check the mechanics one more time.
antiderivative
d/dx (x²) = 2x
differentiating gives back the integrand ✓
order
F(3) − F(0), never F(0) − F(3)
top minus bottom, always
Now you try
Practice problem
The problem
∫₁³ (2x + 1) dx
The antiderivative is x² + x. At the top: 9 + 3 = 12; at the bottom: 1 + 1 = 2; subtract: 12 − 2 = 10. A stops at F(3) and forgets to subtract F(1) — with a bottom limit that is not zero, the subtraction actually moves the answer.