Area between two curves

What you will be able to do

• find the area between any two curves from their meeting points;
• see why the subtraction works when the region lies below the $x$-axis;
• find the area between a parabola and a circle, and between two parabolas.

Any two curves

In the first lesson one curve was a line. The same method works when both are curves. Take a bowl-shaped parabola and an upside-down one, or a parabola and a circle. Cut the region into vertical strips. Each strip has height $f(x) - g(x)$, where $f$ is the upper curve. Add the strips with an integral.

The only new work is in finding the meeting points. Putting $f(x) = g(x)$ may give an equation of higher degree. Factorise it, and keep only the solutions that are real and that fit the picture.

Why the subtraction always works

Area under $f$ minus area under $g$ is correct when both curves are above the $x$-axis. What if the lower curve dips below the axis, or both do? Raise the whole picture by a constant $K$ until both curves are above the axis. Each curve moves up by $K$, so the strip height $(f + K) - (g + K)$ is still $f - g$. The area does not change when you slide the picture up. So the formula $\int (f - g)\,dx$ holds in every position.

A worked example: two parabolas

Find the area between $y = x^2$ and $y = 8 - x^2$.

1. Meeting points. Solve $x^2 = 8 - x^2$. Then $2x^2 = 8$ and $x = \pm 2$.
2. Which is above? At $x = 0$ the downward parabola gives $8$ and the other gives $0$. So $8 - x^2$ is above.
3. Integrate. The area is $\displaystyle\int_{-2}^{2} (8 - 2x^2)\,dx = \left[8x - \dfrac{2x^3}{3}\right]_{-2}^{2} = \dfrac{32}{3} + \dfrac{32}{3} = \dfrac{64}{3}$.

For two parabolas there is a shortcut too. The strip height is a quadratic whose $x^2$ coefficient is the difference of the two coefficients. Here it is $-2$. So the area is $\dfrac{2 \times 4^3}{6} = \dfrac{64}{3}$.

A region below the axis

The line $y = x - 6$ and the parabola $y = x^2 - 4x$ meet where $x^2 - 5x + 6 = 0$, at $x = 2$ and $x = 3$. In between, at $x = 2.5$, the line gives $-3.5$ and the parabola gives $-3.75$. The line is above. Both curves are below the axis, yet the area is

No absolute values are needed. The subtraction takes care of the sign.

A parabola and a circle

Find the area of the part of the disc $x^2 + y^2 \le 2$ that lies above the parabola $y = x^2$.

1. Meeting points. Put $x^2 = y$ into $x^2 + y^2 = 2$. Then $y + y^2 = 2$, so $y = 1$ (the value $y = -2$ is not on the parabola). Then $x = \pm 1$.
2. Upper and lower curves. The upper half of the circle is $y = \sqrt{2 - x^2}$. It is above the parabola between $x = -1$ and $x = 1$.
3. Integrate. The area is $\displaystyle\int_{-1}^{1} \left(\sqrt{2 - x^2} - x^2\right) dx$. Use $\displaystyle\int \sqrt{r^2 - x^2}\,dx = \dfrac{x}{2}\sqrt{r^2 - x^2} + \dfrac{r^2}{2}\sin^{-1}\dfrac{x}{r}$ with $r^2 = 2$.
4. Evaluate. The circle part is $1 + \dfrac{\pi}{2}$ and the parabola part is $\dfrac{2}{3}$. The area is $\dfrac{1}{3} + \dfrac{\pi}{2} \approx 1.9041$.

A curve and its reflection in the line $y = x$ are also easy to meet. For example, $y = \sqrt{x}$ and $y = x^2$ are reflections of each other, and they meet at $x = 0$ and $x = 1$.

Practise with new numbers

Try it in Python

The function area_between adds thin strips for any upper and lower curve. The program uses it for the parabola and circle of the worked example, and then for the region below the axis.

Common mistakes

• Using the $x$-axis as a boundary by habit. The axis is a boundary only when a curve says so.
• Mixing up which curve is above. Check with one test point between the limits.
• Adding areas that should be subtracted. The area under each curve is not the answer. The difference is.
• Keeping false meeting points. A solution that does not lie on both curves, such as $y = -2$ in the circle example, must be dropped.

Remember these

• Area between two curves $= \displaystyle\int_a^b (f - g)\,dx$ with $f$ above $g$ on $[a, b]$.
• The formula is the same above, below or across the $x$-axis.
• For two parabolas, the area is $\dfrac{|a_1 - a_2|\,d^3}{6}$.
• Drop meeting points that are not on both curves.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the area between $y = \sqrt{x}$ and $y = x^2$.

2. Find the area between $y = x^2$ and $y = 2 - x^2$.

3. Find the area between $y = e^x$ and $y = x$ from $x = 0$ to $x = 1$.

4. Find the area between the line $y = -1$ and the parabola $y = x^2 - 4$. The region lies below the axis.

5. Why is the area between two curves not simply the area under one minus a separate area under the other when the region is below the axis?

References

• OpenStax, Calculus Volume 1, Chapter 6, section 6.1 on areas between curves.
• NCERT, Mathematics Part II for Class XII, Chapter 8, the section on the area between two curves.