How to Measure the Area Trapped Between a Parabola and a Line

The area under a curve is made of countless thin vertical strips, each with height $f(x)$ and a tiny width dx. Adding them up is exactly what a definite integral does, so areas bounded by lines, circles, parabolas and far stranger curves can all be found the same way.

This lesson covers areas bounded by simple curves and the axes, other standard curves, the region between two curves, and applied area problems.

**The area bounded by $y=f(x)$, the x-axis and the lines $x=a$ and $x=b$ is ${\int }_a^b|f(x)|dx$, and symmetric curves such as circles and ellipses are handled by finding one quadrant and multiplying by 4.

Worked example (line).** The area under $y=2x+1$ from $x=0$ to $x=3$ is ${\int }_0^3(2x+1)dx=[x^2+x{]}_0^3=12$ square units — the same as the trapezium with parallel sides 1 and 7 and width 3.

Worked example (parabola). The first-quadrant area bounded by $y^2=4x$, the x-axis and $x=4$ is

$${\int }_0^{4}2\sqrt{x}dx={\left[\frac{4}{3}{x}^{3/2}\right]}_0^4=\frac{32}{3}\approx 10.67$$

Worked example (circle). For $x^2+y^2=25$, one quadrant is ${\int }_0^5\sqrt{25-x^2}dx=\frac{25\pi }{4}$, so the whole circle has area $25\pi$.

Worked example (ellipse). For $\frac{x^2}{9}+\frac{y^2}{4}=1$, the area is $4{\int }_0^3\frac{2}{3}\sqrt{9-x^2}dx=\pi \times 3\times 2=6\pi$.

An everyday example. An elliptical flower bed 6 m long and 4 m wide in a park covers $6\pi \approx 18.85$ square metres, exactly as the ellipse integral predicts.

The substance. Integration reproduces familiar formulas — both $\pi r^2$ for a circle and $\pi ab$ for an ellipse come out of the same method.

For any of these curves, integrate between the limits, but split the interval wherever the curve crosses the axis or a modulus changes form, and add the absolute value of each part.

Worked example (polynomial crossing the axis). Find the area between $y=x^2-4$ and the x-axis from $x=0$ to $x=3$. The curve is below the axis on $[0,2]$ and above it on $[2,3]$:

$$\left|{\int }_0^2(x^2-4)dx\right|+{\int }_2^3(x^2-4)dx=\frac{16}{3}+\frac{7}{3}=\frac{23}{3}\approx 7.67$$

Worked example (modulus). The area under $y=|x-2|$ from 0 to 4 is two triangles of area 2 each, so 4 square units.

Worked example (exponential). The area under $y=e^x$ from 0 to 1 is $e-1\approx 1.72$.

Worked example (logarithmic). The area under $y=\log x$ from 1 to e is $[x\log x-x{]}_1^e=0-(-1)=1$.

An everyday example. Measuring a canal bed whose depth dips below a reference level needs every dip counted as a positive depth, just as areas below the x-axis are.

The substance. A single integral across a crossing point understates the area — here ${\int }_0^3(x^2-4)dx=-3$, which is not an area at all.

**The area between $y=f(x)$ and $y=g(x)$ from $x=a$ to $x=b$, where $f(x)\ge g(x)$, is ${\int }_a^b[f(x)-g(x)]dx$, with the limits usually found by solving the two equations together.

Steps:**

- Solve $f(x)=g(x)$ to find the points of intersection
- Decide which curve is on top in each interval
- Integrate the top curve minus the bottom curve

Worked example. Find the area enclosed by $y=x^2$ and $y=x+2$.

- $x^2=x+2$ gives $x=-1$ and $x=2$
- On $[-1,2]$ the line lies above the parabola

$${\int }_{-1}^2(x+2-x^2)dx={\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]}_{-1}^2=\frac{10}{3}+\frac{7}{6}=4.5$$

Worked example 2 (two parabolas). $y^2=4x$ and $x^2=4y$ meet at $(0,0)$ and $(4,4)$, and the area between them is ${\int }_0^4\left(2\sqrt{x}-\frac{x^2}{4}\right)dx=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}$.

An everyday example. A lawn lying between a straight boundary wall and a curved garden path has an area given by exactly this top-minus-bottom integral.

The substance. Top minus bottom works even below the x-axis — the difference $f-g$ stays positive whichever side of the axis the region lies.

For an applied area problem, sketch the region, choose convenient axes, write the boundaries as equations, find the limits from the intersections, integrate top minus bottom, and state the answer in the units of the problem.

Worked example (arch). A tunnel entrance is a parabolic arch 8 m wide and 4 m high. With its base on the x-axis and vertex at $(0,4)$, the arch is $y=4-\frac{x^2}{4}$, and its area is

$${\int }_{-4}^4\left(4-\frac{x^2}{4}\right)dx=2{\left[4x-\frac{x^3}{12}\right]}_0^4=2\left(16-\frac{16}{3}\right)=\frac{64}{3}\approx 21.3{\text{ m}}^2$$

That is two-thirds of the surrounding $8\times 4$ rectangle.

Worked example (triangle). The triangle with vertices $(0,0)$, $(4,0)$ and $(4,3)$ lies under $y=\frac{3x}{4}$, so its area is ${\int }_0^4\frac{3x}{4}dx=6$ square units, matching $\frac{1}{2}\times 4\times 3$.

An everyday example. Estimating the paint for a curved stage backdrop at a school function starts with the area of the region under its curved top edge.

The substance. A sketch prevents most errors — it shows which curve is on top and where the region begins and ends.