Area between a curve and a line

What you will be able to do

• find where a curve and a line meet by solving their two equations together;
• write the area between them as an integral of the upper curve minus the lower curve;
• use the shortcut for a parabola cut by a line.

Land between a stream and a fence

A stream bends like a parabola. A straight fence runs across the bend. How much land lies between the stream and the fence? In the previous course you found the area under one curve. Here the region has a curve on top and another curve below it.

The idea

Cut the region into thin vertical strips. A strip at position $x$ has width $dx$. Its height is the distance from the lower curve to the upper curve. Let $f$ be the upper curve and $g$ the lower curve. Then

The numbers $a$ and $b$ are the $x$-values of the two places where the curves meet. This is the area under $f$ minus the area under $g$, done in one step.

Step 1: sketch, then find the meeting points

Always draw the picture first. For a parabola, find which way it opens and where its lowest or highest point is. Mark the two curves on one set of axes. The picture shows which curve is above, and it shows how many regions there are.

Then solve $f(x) = g(x)$. The solutions are the limits $a$ and $b$. To find which curve is above, put one $x$ between $a$ and $b$ into both formulas and compare.

Books word this differently. NCERT says "the area of the region bounded by" the curves. A-Level and IB often say "find the area enclosed by". AP says "the region bounded by". All of them mean the finite region that the curves close off. A sketch is expected in every one of them.

A worked example

Find the area between the parabola $y = x^2$ and the line $y = x + 2$.

1. Meeting points. Solve $x^2 = x + 2$. Then $x^2 - x - 2 = 0$, which factorises as $(x - 2)(x + 1) = 0$. So $x = -1$ or $x = 2$.
2. Which is above? At $x = 0$ the line gives $2$ and the parabola gives $0$. The line is above.
3. Integrate. The area is $\displaystyle\int_{-1}^{2} (x + 2 - x^2)\,dx = \left[\dfrac{x^2}{2} + 2x - \dfrac{x^3}{3}\right]_{-1}^{2}$.
4. Evaluate. At $x = 2$ the bracket is $2 + 4 - \dfrac{8}{3} = \dfrac{10}{3}$. At $x = -1$ it is $\dfrac{1}{2} - 2 + \dfrac{1}{3} = -\dfrac{7}{6}$. The area is $\dfrac{10}{3} + \dfrac{7}{6} = \dfrac{9}{2}$.

A shortcut for a parabola cut by a line

Suppose a line meets the parabola $y = ax^2 + bx + c$ at $x_1$ and $x_2$. The strip height is the difference of the two formulas, which is a quadratic that is zero at $x_1$ and $x_2$. So it equals $a(x - x_1)(x - x_2)$ up to sign. The integral of $(x - x_1)(x - x_2)$ from $x_1$ to $x_2$ is $-\dfrac{(x_2 - x_1)^3}{6}$. Therefore

Check it on the example. Here $a = 1$ and $x_2 - x_1 = 3$, so the area is $\dfrac{27}{6} = \dfrac{9}{2}$. The shortcut gives the same answer as the integral.

The shortcut needs a parabola and a line. For any other pair of curves, integrate.

Practise with new numbers

Try it in Python

The program finds the meeting points of $y = x^2$ and $y = 3x + 10$. Then it adds up thin strips and compares the total with the shortcut.

Common mistakes

• Guessing the limits. The limits are the solutions of $f(x) = g(x)$. Read them off the equations, not off the picture.
• Subtracting the wrong way. If you take lower minus upper, the answer is negative. An area is never negative, so test a point and swap.
• Taking the area under the upper curve only. The region between the curves is smaller. Subtract the lower curve.
• Using the shortcut for other curves. The factor $\dfrac{1}{6}$ belongs to a parabola cut by a line.

Remember these

• Area between curves $= \displaystyle\int_a^b (\text{upper} - \text{lower})\,dx$, where $a$ and $b$ are the meeting points.
• Sketch first. Solve $f(x) = g(x)$ for the limits.
• Test one $x$ between the limits to see which curve is above.
• A parabola cut by a line has area $\dfrac{|a|\,d^3}{6}$.

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 $y = x^2$ and $y = 2x + 8$.

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

3. Find the area between $y = 6x - x^2$ and $y = x$.

4. A student integrates $g - f$ instead of $f - g$, where $f$ is above $g$. What does she get for the example $y = x + 2$ above $y = x^2$?

5. Find the area between $y = x^2$ and the horizontal line $y = 4$.

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.