IB Maths AI SLIntegrationPaper 1 & 2definite integrals~6 min read
Finding Areas Using a GDC
The area trapped between a curve and the x-axis is found exactly by a definite integral. Unlike the trapezoidal rule, this gives the true area, not an estimate. This note covers what a definite integral is, how it gives an area, and how to set up the limits.
📘 What you need to know
The area under a curve is the region bounded by y = f(x), the x-axis, and the vertical lines x = a and x = b.
This exact area is a definite integral, ∫abf(x) dx — a is the lower limit, b the upper limit.
The Fundamental Theorem of Calculus: ∫abf(x) dx = F(b) − F(a).
The constant of integration is not needed for a definite integral — it cancels when you subtract.
The limits come from vertical lines, from the y-axis (x = 0), or from roots of f(x) = 0.
A GDC evaluates definite integrals directly — but may give a decimal, so give an exact answer when one is asked for.
Area under a curve
The phrase area under a curve means a very specific region: the area bounded by the graph of y = f(x), the x-axis, and two vertical lines x = a and x = b. Its exact value is a definite integral.
Area under a curveA = ∫abf(x) dxthe exact area bounded by the curve, the x-axis, and the lines x = a and x = b
The shaded region is bounded by the curve, the x-axis, and the lines x = a and x = b. Its exact area is the definite integral from a to b.
Unlike the trapezoidal rule, which only estimates, a definite integral gives the exact area. The integral ∫abf(x) dx has an integrandf(x) and two limits: a is the lower limit, b the upper limit.
Evaluating a definite integral
A definite integral is evaluated using the Fundamental Theorem of Calculus. Integrate the function to get an antiderivative F(x), then subtract its value at the lower limit from its value at the upper limit.
The Fundamental Theorem of Calculus
∫abf(x) dx = F(b) − F(a)
F is an antiderivative of f · the + c is not needed — it cancels in the subtraction
Notice there is no “+ c“. A constant of integration would appear with both F(b) and F(a), and subtracting cancels it — so definite integrals never need it. A GDC can evaluate the integral directly, though it may return a decimal; give an exact answer whenever one is asked for.
Finding the limits
The curve and the x-axis are obvious boundaries; the work is usually in finding the limits a and b. They come from one of three places.
If the region is bounded by vertical lines, the limits are read straight from their equations. If the y-axis is a boundary, that limit is x = 0. And if the region runs up to where the curve meets the x-axis, the limit is a root — solve f(x) = 0 to find it.
🧠Recipe — finding an area under a curve
Sketch and shade the region whose area you want — plot the curve on your GDC if it helps.
Identify the limitsa and b — from vertical lines, the y-axis (x = 0), or roots of f(x) = 0.
Set up the definite integralA = ∫abf(x) dx.
Evaluate it — by F(b) − F(a), or directly on your GDC.
State the answer with units (square units), exact if an exact form is asked for.
Worked examples
WE 1
A basic definite integral
Evaluate ∫13 2x dx.
integrate to find an antiderivative: F(x) = x2apply F(b) − F(a) with a = 1, b = 3F(3) − F(1) = (3)2 − (1)2 = 9 − 1∫13 2x dx = 8no “+ c” — it would simply cancel in the subtraction.
WE 2
A polynomial integrand
Evaluate ∫02 (3x2 + 1) dx.
antiderivative: F(x) = x3 + xF(2) − F(0) = (8 + 2) − (0 + 0)= 10 − 0∫02 (3x2+1) dx = 10integrate term by term, then substitute the two limits.
WE 3
The y-axis as a limit
Find the area bounded by the curve y = 4 − x2, the x-axis and the y-axis, in the first quadrant.
the y-axis gives the lower limit: x = 0the curve meets the x-axis where 4 − x2 = 0 ⇒ x = 2area = ∫02 (4 − x2) dx, F(x) = 4x − x3/3F(2) − F(0) = (8 − 8/3) − 0area = 16/3 square unitsthe y-axis is the line x = 0; the other limit is a root of f(x) = 0.
WE 4
Both limits are roots
Find the area enclosed between the curve y = x(6 − x) and the x-axis.
find where the curve meets the x-axisx(6 − x) = 0 ⇒ x = 0 and x = 6 (the limits)expand: y = 6x − x2, F(x) = 3x2 − x3/3F(6) − F(0) = (108 − 72) − 0area = 36 square unitswhen both boundaries lie on the x-axis, both limits are roots of f(x) = 0.
WE 5
Vertical lines as limits
A region is bounded by y = x3 + 2, the x-axis, and the lines x = 1 and x = 3. Write down a definite integral for its area, then evaluate it.
the vertical lines give the limits directly: a = 1, b = 3area = ∫13 (x3 + 2) dx, F(x) = x4/4 + 2xF(3) − F(1) = (81/4 + 6) − (1/4 + 2)= 105/4 − 9/4 = 96/4area = 24 square unitsvertical-line boundaries give the limits with no extra work.
WE 6
Full question: an enclosed area
A curve has equation y = 12 − 3x2. (a) Find where the curve crosses the x-axis. (b) Write down a definite integral for the area enclosed between the curve and the x-axis. (c) Find this area exactly.
(a) the curve crosses the x-axis where 12 − 3x2 = 03x2 = 12 ⇒ x2 = 4 ⇒ x = −2 and x = 2(b) these roots are the limits of the areaarea = ∫−22 (12 − 3x2) dx(c) F(x) = 12x − x3F(2) − F(−2) = (24 − 8) − (−24 + 8) = 16 − (−16)(a) x = ±2 · (b) ∫−22(12−3x2) dx · (c) area = 32 square unitsa GDC gives 32 directly — but for a messier curve it may show a decimal, so an exact answer needs the working.
💡 Top tips
A definite integral gives an exact area; the trapezoidal rule only estimates one.
Evaluate with F(b) − F(a) — upper limit value minus lower limit value.
Definite integrals need no “+ c“ — it cancels in the subtraction.
Limits come from vertical lines, the y-axis (x = 0), or roots of f(x) = 0.
A GDC evaluates definite integrals directly, but may give a decimal — give an exact answer when asked.
âš Common mistakes
Adding “+ c” to a definite integral — it is never needed.
Subtracting the wrong way round — it is F(b) − F(a), upper minus lower.
Forgetting that the y-axis is the line x = 0 when it bounds the region.
Not solving f(x) = 0 to find a limit that lies on the x-axis.
Giving a rounded GDC decimal when the question asks for an exact answer.
That completes Integration. Look back at the arc: the trapezoidal rule estimates an area; antidifferentiation reverses the derivative; the power rule and the constant of integration build indefinite integrals; and the definite integral — F(b) − F(a) — finally delivers an exact area. Differentiation and integration are two directions of one idea: master moving both ways and the calculus of AI SL is yours.
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