Geometry gives formulas for triangles, rectangles and trapezia, but not for a region with a curved edge. Integration does: slice the region into thin strips, each almost a rectangle, and add them up. That one idea gives the area of a circle, an ellipse, and any region under a curve, including parts that dip below the axis.
Cut the region under a curve into very many thin vertical strips. Add up their areas and you have the whole area.
A strip at x has height y and width dx. Its area is called the elementary area.
Add the strips from x = a to x = b. This is the area between y = f(x), the x-axis and the lines x = a and x = b. Work it out as F(b) − F(a) using an antiderivative (Chapter 7).
This formula is for the region between the curve and the x-axis. If the curve dips below the axis, the integral of that part is negative: see Part 3.
When a region sits against the y-axis, slice it sideways instead, into horizontal strips.
A strip at height y has length x and thickness dy.
This is the region bounded by a curve x = g(y), the y-axis and the lines y = c and y = d.
The curve’s equation must give x in terms of y before you integrate.
An area is only a size, so it is never negative. Below the x-axis, take the absolute value.
If f(x) < 0 on [a, b], the curve is below the x-axis and the integral comes out negative. The area is its size.
Split at the points where the curve crosses the axis, then add the sizes. Never let the pieces cancel.
y = 3x + 2, x-axis, x = −1 to x = 1.
y = cos x from 0 to 2π.
Circles and ellipses are symmetric about both axes. Find one quarter of the area, then multiply by four.
Both curves are symmetric about both axes, so the four quarters have equal areas.
It is the area of a quarter circle of radius a. So the circle has area πa² and the ellipse πab.
| Curve | y in the first quadrant |
|---|---|
| x² + y² = a² | √(a² − x²) |
| x²a² + y²b² = 1 | ba √(a² − x²) |
Area of the circle x² + y² = a².
Area of the ellipse x²/a² + y²/b² = 1.
Every MCQ from Exercise 8.1 and the miscellaneous exercise, plus four concept checks. Tap an option to see the reasoning.
Area between y = f(x), the x-axis and x = a, x = b (b > a).
Area between x = φ(y), the y-axis and y = c, y = d.
Below the axis the integral is negative: take |∫|. If the curve crosses the axis, split and add the sizes.
Each is 4 × its first-quadrant quarter.
The "method of exhaustion": fill a curved region with more and more polygons until nothing is left over.
Used exhaustion to find the area of a circle and of a parabolic segment (2/3 of its rectangle).
Newton's fluxions and Leibniz's "calculus summatorius": areas as sums of infinitely thin strips, written with the ∫ sign.
Made the idea of a limit precise, so that "infinitely thin strips" became rigorous mathematics.
Each example is an animated story on one graph. The region fills in, thin strips slice it, one highlighted strip shows its area dA = y dx (or x dy), and the strips are added up by an integral. Circles and ellipses are done twice, with vertical and then horizontal strips. Regions below the x-axis turn red: their integral is negative, so we add its size. The last step always measures the shaded area by adding 400 strips and compares it with the answer. Press Next step (or Play). Under the controls each step is explained in detail, with links to the exact basics it uses; after reading a basic, press ← Back to return to the same step.
Use symmetry to compute one quarter, slice it into strips, and add them with ∫√(a² − x²) dx.
Split where the curve crosses the x-axis and add the sizes of the pieces.
A field along a winding river, an oval lawn, the paint on a wave-shaped sign, a lake surveyed from its shore, and a car that drives forward and then reverses: each is a region with a curved edge, measured by adding thin strips.
A farmer's field is bounded by a straight road (the x-axis), two straight fences, and the winding river bank y = 20 + 6 sin(x/8) + x/10 metres. A surveyor measures the depth of the field every few metres and adds up strips. More measurements, better answer; the integral is the limit.
An oval lawn is an ellipse with half-length a and half-width b metres. Its area is πab (Example 2). Turf comes in rolls of 1 m², so the gardener orders πab rolls, rounded up.
A sign has a wavy stripe between the line y = 0 and the curve y = A sin x, painted both above and below. The paint needed is the area, not the integral: over each full wave the integral is 0, but the area is 4A.
A surveyor walks north along a straight path and, every few metres, measures how wide the lake is from west to east. Each measurement is a horizontal strip of width x(y) and thickness dy; the area is ∫ x dy.
A car's velocity is v(t) = 4 sin(πt/6) m/s for 12 seconds: forward for 6 s, then backward. The signed area under v is the displacement (where the car ends up); the actual area is the distance driven, which is what the odometer shows.
Decide first, then tap each card.
All six questions that ask you to find or evaluate an area have an animated walkthrough here. The four multiple-choice questions are in the Lesson’s self-check, and they are also worked out in their own player below. A last player adds five practice questions beyond the book. Pick a question from the chips above each player, then press Play or step through with Next.
Read off a and b, compute one quarter with vertical strips, and multiply by 4. (Q3–Q4 are in the quiz and in the MCQ player.)
Integrate where the curve is above the axis; split and add sizes where it dips below. (Q4–Q5 are in the quiz and in the MCQ player.)
Exercise 8.1 Q3–Q4 and Miscellaneous Q4–Q5, animated. The trap options are usually the signed integral, where pieces below the axis cancel.
Not from the textbook: extra questions of the same kinds, for practice.
This chapter uses the definite integral from Chapter 7 to measure real areas. For that you need seven ideas, each explained from its roots: what a definite integral is, why an area is a sum of thin strips, why area is never negative, how to read a curve from its equation, the symmetry of circles and ellipses, where ∫√(a² − x²) dx comes from, and how to treat x as a function of y. Every example and exercise links here exactly where you need it, and a button takes you straight back to the problem.
A definite integral is a number. Chapter 7 gave a shortcut for finding it.
In Chapter 7 the definite integral was defined as the limit of sums of thin strips. The Fundamental Theorem gives this shortcut: find any antiderivative F (so F′ = f). The constant C cancels in the subtraction, so it is left out.
Here the antiderivative is F(x) = x³/3.
The full story, with every rule and technique of integration, is on the Chapter 7 page.
A region under a curve is not a rectangle, but a very thin vertical slice of it almost is.
A rectangle’s area is height × width. The slice has height y = f(x) and a tiny width dx, so this is its elementary area.
The whole area is the sum of all the slices from x = a to x = b. As the slices get thinner, the sum becomes the integral. The ∫ sign is a stretched S, for “sum”.
Below the x-axis the height y is negative, so every strip y dx is negative, and so is the integral. But a region has a positive size.
For a part below the axis, take the absolute value.
Split the integral at the crossing points, find each piece, and add their sizes. The plain integral lets red and green cancel; the area never does.
To integrate we need y as a formula in x. Knowing where a curve meets the axes tells you the limits.
Already solved for y.
From y² = a² − x². Take + for the upper half and − for the lower half.
It crosses the axes at (±a, 0) and (0, ±b).
It opens to the right.
Circles and ellipses have four equal quarters. So find one quarter and multiply by four.
If replacing y by −y does not change a curve’s equation, the curve is its own mirror image in the x-axis. If replacing x by −x does not change it, it is symmetric about the y-axis.
They pass both tests (only x² and y² appear), so the four quarters have equal areas. In the first quadrant x and y are both positive and the + root applies.
The formula looks complicated, but the picture explains it: a triangle plus a sector.
This is the area under the arc from 0 to x.
Base x and height y = √(a² − x²).
A sector of the circle with angle θ at the centre.
The triangle vanishes and θ = π/2, giving the quarter circle.
If a region is bounded by a curve and the y-axis, slice it horizontally. It is the same idea as vertical strips, with x and y swapped.
A strip at height y has length x = g(y) and thickness dy.
Add the strips from y = c to y = d.
You get x as a function of y by solving the equation for x.
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