Math II

Area between curves

Definite integrals give the area of a region bounded by curves. The key is knowing which graph is on top. Every problem comes with a figure.

Basic, Standard: Grade 11 · Term 3 / Advanced: Grade 12 · Exam prep

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How to find an area

Area between two curves

If $f(x) \ge g(x)$ on $a \le x \le b$ , the area $S$ bounded by $y = f(x)$ , $y = g(x)$ and the lines $x = a$ , $x = b$ is

S = \int_a^b \{f(x) - g(x)\}\,dx

Remember: integrate (upper) − (lower). The area of a part below the $x$ -axis (where $f(x) \le 0$ ) is $\displaystyle -\int_a^b f(x)\,dx$ .

Step by step

Sketch the graphs (a rough sketch is enough).
Find the $x$ -coordinates of the intersection points to get the limits.
Check which graph is on top. If they swap, split the interval there.
Integrate (upper) − (lower).

Faster with the 1/6 formula

For the area between a parabola and a line, or two parabolas, the difference has the form $a(x - \alpha)(x - \beta)$ , where $\alpha < \beta$ are the intersection points. Using

\int_\alpha^\beta (x-\alpha)(x-\beta)\,dx = -\frac{1}{6}(\beta-\alpha)^3

the area is $S = \dfrac{|a|}{6}(\beta - \alpha)^3$ . You only need $\beta - \alpha$ , even when the roots are not nice numbers. For the roots of $ax^2 + bx + c = 0$ , $\beta - \alpha = \dfrac{\sqrt{b^2 - 4ac}}{|a|}$ .

A parabola and its tangent (the 1/3 formula)

The difference between the parabola $y = ax^2 + bx + c$ and its tangent line at $x = t$ is $a(x - t)^2$ . So the area bounded by the parabola, the tangent and the line $x = s$ is

S = \left|\int_t^s a(x-t)^2\,dx\right| = \frac{|a|}{3}|s - t|^3

Because the graphs touch at the point of tangency, the difference is a perfect square.

Common mistakes

Upside down: integrating (lower) − (upper) gives a negative number. Areas are always positive.
Not splitting the interval: if the graphs swap (as with cubics), one integral lets positive and negative parts cancel.
The coefficient in the 1/6 formula: if the coefficient of $x^2$ is not 1, multiply by $|a|$ .

Q1BasicGr 11 · Term 3Area between curves

Find the area $S$ of the region bounded by the curve $y = x^{2} + 2$ , the $x$ -axis and the lines $x = -1$ and $x = 3$ .

y = x^{2} + 2

Answer

S = \frac{52}{3}

Solution

Since $x^{2} + 2 > 0$ ,
$\begin{aligned}S &= \int_{-1}^{3} (x^{2} + 2)\,dx \\ &= \left[\frac{1}{3}x^{3} + 2x\right]_{-1}^{3} \\ &= (9 + 6) - \left(-\frac{1}{3} - 2\right) \\ &= \frac{52}{3}\end{aligned}$

Q2StandardGr 11 · Term 3Area between curves

Find the area $S$ of the region bounded by the parabola $y = x^{2} + x - 2$ and the line $y = 0$ .

y = x^{2} + x - 2

y = 0

Answer

S = \frac{9}{2}

Solution

The $x$ -coordinates of the intersection points satisfy
$x^{2} + x - 2 = 0 \quad\text{so}\quad x^{2} + x - 2 = 0$
$(x + 2)(x - 1) = 0$ , so $x = -2,\ 1$ .
On $-2 \le x \le 1$ the line lies above the parabola, so use the formula $\displaystyle\int_\alpha^\beta (x-\alpha)(x-\beta)\,dx = -\frac{1}{6}(\beta-\alpha)^3$ .
$\begin{aligned}S &= \int_{-2}^{1} (-x^{2} - x + 2)\,dx \\ &= \frac{1}{6}\{1 - (-2)\}^{3} \\ &= \frac{9}{2}\end{aligned}$

Q3AdvancedExam prepArea between curves

Let $\ell$ be the tangent line to the parabola $y = x^{2} - 2x - 3$ at the point $(-3,\ 12)$ . Find the area $S$ of the region bounded by the parabola, the tangent line $\ell$ and the $y$ -axis.

y = x^{2} - 2x - 3

\ell : y = -8x - 12

Answer

S = 9

Solution

Since $y' = 2x - 2$ , the slope of the tangent at $x = -3$ is $-8$ . So the equation of $\ell$ is
$y - 12 = -8(x + 3) \quad\text{i.e.}\quad y = -8x - 12$
The parabola lies above the tangent line, and (parabola) − (tangent) $= x^{2} + 6x + 9 = (x + 3)^{2}$ , so
$\begin{aligned}S &= \int_{-3}^{0} (x + 3)^{2}\,dx \\ &= \left[\frac{1}{3}(x + 3)^{3}\right]_{-3}^{0} \\ &= 9\end{aligned}$

Q1BasicGr 11 · Term 3Area between curves

Find the area $S$ of the region bounded by the parabola $y = 2x^{2} - 10x + 12$ and the $x$ -axis.

y = 2x^{2} - 10x + 12

Q2StandardGr 11 · Term 3Area between curves

Find the area $S$ of the region bounded by the parabola $y = x^{2} + x - 2$ and the line $y = -2x - 2$ .

y = x^{2} + x - 2

y = -2x - 2

Q3AdvancedExam prepArea between curves

Let $\ell$ be the tangent line to the parabola $y = x^{2} - 1$ at the point $(-1,\ 0)$ . Find the area $S$ of the region bounded by the parabola, the tangent line $\ell$ and the $y$ -axis.

y = x^{2} - 1

\ell : y = -2x - 2

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