Part 5 · Chapter 27

Area Under Curves

From signed integral to true geometric area — measuring the region a curve traps against the axis or against another curve, choosing vertical or horizontal slices, and letting symmetry do half the work

Fundamentals of Mathematics Prof. Mithun Mondal Reading time ≈ 48 min
i What you'll learn
  • The area trapped between a curve and the x-axis as a definite integral of vertical strips.
  • The difference between signed and geometric area when a curve dips below the axis.
  • The area between two curves as the integral of top minus bottom.
  • Integrating with respect to \(y\) using horizontal strips, and finding limits from intersections.
  • The standard areas of a circle, ellipse and parabola.
  • Using symmetry, splitting composite regions, and the parametric area formula.
Section 27-1

Area Under a Curve

The definite integral was built as a limit of thin rectangles, so when \(f(x)\ge 0\) it is the area trapped between the curve, the \(x\)-axis, and the vertical lines \(x=a\) and \(x=b\). Each strip has height \(f(x)\) and width \(dx\); the integral sums them.

a b f(x) dx
Each strip contributes \(f(x)\,dx\); the integral adds them from \(a\) to \(b\)
📐
Area beneath a curve
\(\displaystyle A=\int_a^b f(x)\,dx,\qquad f(x)\ge 0\text{ on }[a,b]\)

This is the workhorse of the chapter. The integrand is the height of the region at each \(x\), and \(dx\) is the width of a slice. The same idea generalises to every region in the chapter — only the height changes.

Section 27-2

When the Curve Dips Below the Axis

Below the axis the height \(f(x)\) is negative, so the integral counts that area as negative. For the geometric area — the actual amount of ink — you must find where the curve crosses the axis, integrate each piece separately, and add the magnitudes.

+ − a c b
The lobe below the axis carries a minus sign in the raw integral
±
Signed versus geometric area
\(\displaystyle A_{\text{geom}}=\int_a^b |f(x)|\,dx=\int_a^c f\,dx-\int_c^b f\,dx\)

Here \(c\) is the \(x\)-intercept where the curve changes sign. The raw integral \(\int_a^b f\,dx\) gives the net signed area, which can be smaller than the true area, or even zero. When a problem asks for "the area," it almost always wants the geometric value.

Section 27-3

Area Between Two Curves

The region trapped between two curves has, at each \(x\), a height equal to the upper curve minus the lower one. Summing those strips gives the enclosed area — and because we subtract, it does not matter whether the curves sit above or below the axis.

g−f g(x) f(x) a b
The strip runs from the lower curve up to the upper one: height \(g-f\)
↕
Top minus bottom
\(\displaystyle A=\int_a^b \big[g(x)-f(x)\big]\,dx,\qquad g\ge f\text{ on }[a,b]\)

The limits \(a\) and \(b\) are the \(x\)-coordinates where the curves meet. Setting Section 27-1 as the special case \(f\equiv 0\) recovers the area under a single curve.

! Sketch first — and watch for a swap

Always draw the region before writing the integral. The upper curve can change from one to the other at an intersection inside \([a,b]\); when that happens, split the integral at the crossing and use the correct "top minus bottom" on each piece. Blindly integrating \(g-f\) across a swap gives a wrong, partly-cancelled answer.

Section 27-4

Integrating Along the y-axis

When a boundary is given more naturally as \(x=h(y)\) — a sideways parabola, or a region measured from the \(y\)-axis — it is far cleaner to slice the region into horizontal strips of width \(dy\) and length \(x\), and integrate up the \(y\)-axis.

c d x=h(y) x
Horizontal strips: length \(x=h(y)\), width \(dy\), summed from \(c\) to \(d\)
↔
Horizontal strips
\(\displaystyle A=\int_c^d x\,dy=\int_c^d h(y)\,dy\)

Everything is the mirror image of the vertical case: the roles of \(x\) and \(y\) simply swap, and the limits \(c,d\) are now \(y\)-values. For a region between two side-by-side curves, the strip length is \((\text{right})-(\text{left})\).

Section 27-5

Locating the Limits

The hardest part of most area problems is not the integration but pinning down the limits. They come from the points of intersection of the boundaries — solve the curves simultaneously, and the solutions are your \(a\) and \(b\) (or \(c\) and \(d\)).

✕
Limits from intersections
\(\displaystyle f(x)=g(x)\ \Rightarrow\ x=a,\,b\)

To find where a curve meets a line or another curve, equate them and solve. A quadratic gives two crossings, which become the limits; more crossings mean the region splits into pieces. Always confirm with a sketch which curve is on top between consecutive crossings.

Section 27-6

Standard Regions

A handful of areas recur so often that their formulas are worth knowing outright — and each one falls straight out of the integrals from Chapter 25, usually via \(\int\sqrt{a^2-x^2}\,dx\) and a dose of symmetry.

a b ¼ area
Integrate one quarter, then multiply by four: ellipse area \(=\pi a b\)
⬭
Circle, ellipse, parabola
\(\displaystyle \text{circle } \pi r^{2};\quad \text{ellipse } \pi a b;\quad \text{parabolic segment } \tfrac{2}{3}(\text{base})(\text{height})\)

The circle is the special case \(a=b=r\) of the ellipse. The parabolic-segment result — Archimedes' theorem — says the area cut off by a chord is two-thirds of the enclosing rectangle. Each is a quick integral once symmetry trims it to a single quadrant.

Section 27-7

Exploiting Symmetry

If a region is symmetric about an axis or the origin, compute the area of one representative piece and multiply. This not only saves work but also sidesteps the sign trouble of Section 27-2, since you integrate over a region where the height keeps one sign.

−a a ×2 axis of symmetry
Symmetric about the \(y\)-axis: find half the area and double it
🪞
Multiply one piece by the symmetry count
\(\displaystyle A=2\int_0^a f(x)\,dx\ \ (\text{even, about }y\text{-axis});\qquad A=4\times(\text{one quadrant})\)

A figure symmetric about the \(y\)-axis needs only the right half; one symmetric about both axes, like a circle or ellipse, needs only a single quadrant. Identify the symmetry first — it is often the difference between a one-line answer and a page of algebra.

Section 27-8

Composite & Parametric Regions

Some regions are bounded by three or more curves, with the "top" or "bottom" changing along the way. The cure is to split the region at every such transition and add the pieces. And when a boundary is given parametrically, the area formula adapts in a natural way.

🧩
Splitting and the parametric form
\(\displaystyle A=\sum_k\int_{a_k}^{a_{k+1}}(\text{top}-\text{bottom})\,dx;\qquad A=\int_{t_1}^{t_2} y\,\frac{dx}{dt}\,dt\)

Break a composite region at each corner or crossing so that a single "top minus bottom" rule holds on every sub-interval, then sum. For a curve traced by \(x=x(t),\ y=y(t)\), replace \(dx\) by \(x'(t)\,dt\) — exactly how the area of one arch of a cycloid is found.

Worked Examples

Putting It to Work

1 Area under a parabola

Problem. Find the area bounded by \(y=x^{2}\), the \(x\)-axis, and \(x=0,\ x=2\).

Solution. The curve is above the axis, so integrate directly:

Working
\[ A=\int_0^{2} x^{2}\,dx=\Big[\tfrac{x^{3}}{3}\Big]_0^{2}=\frac{8}{3} \]
2 A region split at the axis

Problem. Find the geometric area between \(y=x^{3}\), the \(x\)-axis, and \(x=-1,\ x=1\).

Solution. The curve is below the axis on \([-1,0]\) and above on \([0,1]\), so add magnitudes:

Working
\[ A=\left|\int_{-1}^{0} x^{3}\,dx\right|+\int_0^{1} x^{3}\,dx=\frac14+\frac14=\frac12 \]
3 Between two curves

Problem. Find the area enclosed between \(y=x\) and \(y=x^{2}\).

Solution. They meet where \(x=x^{2}\), i.e. \(x=0,1\); on \((0,1)\) the line is on top:

Working
\[ A=\int_0^{1}(x-x^{2})\,dx=\Big[\tfrac{x^{2}}{2}-\tfrac{x^{3}}{3}\Big]_0^{1}=\frac12-\frac13=\frac16 \]
4 Horizontal strips

Problem. Find the area enclosed by the parabola \(x=y^{2}\) and the line \(x=4\).

Solution. They meet at \(y=\pm2\); a horizontal strip runs from \(x=y^{2}\) to \(x=4\), and the region is symmetric in \(y\):

Working
\[ A=\int_{-2}^{2}(4-y^{2})\,dy=2\int_0^{2}(4-y^{2})\,dy=2\Big[4y-\tfrac{y^{3}}{3}\Big]_0^{2}=\frac{32}{3} \]
5 Area of an ellipse

Problem. Find the area enclosed by \(\dfrac{x^{2}}{9}+\dfrac{y^{2}}{4}=1\).

Solution. By four-fold symmetry, integrate the first-quadrant arc \(y=\tfrac23\sqrt{9-x^{2}}\) and use \(\int_0^{3}\sqrt{9-x^{2}}\,dx=\tfrac{9\pi}{4}\):

Working
\[ A=4\int_0^{3}\frac23\sqrt{9-x^{2}}\,dx=\frac{8}{3}\cdot\frac{9\pi}{4}=6\pi\quad(=\pi a b) \]
6 Using symmetry

Problem. Find the area bounded by \(y=9-x^{2}\) and the \(x\)-axis.

Solution. The parabola meets the axis at \(x=\pm3\) and is symmetric about the \(y\)-axis, so double the right half:

Working
\[ A=2\int_0^{3}(9-x^{2})\,dx=2\Big[9x-\tfrac{x^{3}}{3}\Big]_0^{3}=2(27-9)=36 \]
Review

Chapter Summary

✓ Under a curve

\(A=\int_a^b f\,dx\) for \(f\ge 0\); vertical strips of height \(f\).

✓ Below the axis

Split at the intercepts and add magnitudes: \(\int|f|\,dx\).

✓ Between curves

\(A=\int_a^b(\text{top}-\text{bottom})\,dx\); sketch first.

✓ Along the y-axis

\(A=\int_c^d x\,dy\) with horizontal strips.

✓ Limits

Solve the curves simultaneously to find the crossings.

✓ Symmetry & standards

Circle \(\pi r^2\), ellipse \(\pi a b\); one piece × count.

Practice

Problems

Sketch the region first, decide between vertical and horizontal strips, find the limits from the intersections, and only then integrate. Difficulty rises down the list.

  1. Find the area under \(y=x^{2}\) from \(x=0\) to \(x=3\).
  2. Find the area under \(y=2x\) from \(x=1\) to \(x=4\).
  3. Find the area bounded by \(y=\sin x\) and the \(x\)-axis on \([0,\pi]\).
  4. Find the geometric area between \(y=x^{3}\), the \(x\)-axis, \(x=-2\) and \(x=2\).
  5. Find the area between \(y=x^{2}\) and \(y=2x\).
  6. Find the area enclosed by \(x=y^{2}\) and \(x=9\) using horizontal strips.
  7. Find the area enclosed by the circle \(x^{2}+y^{2}=16\).
  8. Find the area enclosed by the ellipse \(\dfrac{x^{2}}{16}+\dfrac{y^{2}}{9}=1\).
  9. Find the area bounded by \(y=4-x^{2}\) and the \(x\)-axis.
  10. Find the area between the parabolas \(y^{2}=4x\) and \(x^{2}=4y\).
  11. Find the area bounded by the parabola \(y=x^{2}\) and the line \(y=x+2\).
  12. Find the area bounded by \(y^{2}=4x\) and the line \(x=1\) using horizontal strips.
  13. Find the area between \(y=\sin x\) and \(y=\cos x\) on \(\big[0,\tfrac\pi4\big]\).
Tip: nearly every area problem follows the same four steps — sketch, choose the slice direction, find the limits, integrate. Reach for horizontal strips whenever the boundary is written as \(x=h(y)\) or the curve is a sideways parabola; it often turns an ugly split into a single clean integral. Test for symmetry before grinding through algebra, and when a curve crosses the axis inside the interval, remember that "the area" means the geometric area, so split at the intercepts and add magnitudes.