Area under a curve: applications of integration

Three places where an area matters: the opening of a covered walkway, the end cross-section of a plunge pool, and a lawn between a fence and a stream. The pictures come first; the working in each solution comes from the same numbers.

$y = \dfrac{9}{2} - \dfrac{x^{2}}{2}$$6\text{ m wide}$$x = -3$$x = 3$$O$$\text{ground}$

Maths In Real Life 1

The opening of a covered walkway

Covered walkways in housing estates often end in a curved arch. The area of the opening tells the builder how much of the end wall has to be cut out or covered.

The end of a covered walkway is an arch. Take the ground as the $x$-axis and the middle of the opening on the ground as the origin, with $x$ and $y$ in metres. The top edge of the arch is the curve $y = - \frac{x^{2}}{2} + \frac{9}{2}$ for $-3 \le x \le 3$.

• (a)Find the area of the opening of the arch.

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Colour keythe evaluated area or value

(a)

$\displaystyle \dfrac{-3 + 3}{2} = 0$

The middle of the piece is a safe test value: it cannot be a root.

$\displaystyle - \frac{\left(0\right)^{2}}{2} + \frac{9}{2} = \frac{9}{2}$

The curve's height at the test value: positive, so this piece lies above the $x$-axis.

$\displaystyle \int_{-3}^{3} \left(- \frac{x^{2}}{2} + \frac{9}{2}\right)\,dx = \left[- \frac{x^{3}}{6} + \frac{9 x}{2}\right]_{-3}^{3}$

Integrate with the usual rules but leave out $c$: it cancels in a definite integral.

$\displaystyle = \left(- \frac{\left(3\right)^{3}}{6} + \frac{9 \times \left(3\right)}{2}\right) - \left(- \frac{\left(-3\right)^{3}}{6} + \frac{9 \times \left(-3\right)}{2}\right)$

Substitute the upper limit $3$, then the lower limit $-3$, and subtract (upper minus lower).

$\displaystyle = \textcolor{#010106}{18}$

Evaluate.

The curve is above the axis, so the area is the integral itself.

Answer

(a) $18$ m²

$y = \frac{x^{2}}{4} - 2x$$(0,\ 0)$$(8,\ 0)$$(4,\ -4)$$\text{water surface}$$\begin{array}{c}\text{6 m wide} \\ \text{(into the page)}\end{array}$$\text{end view}$

Maths In Real Life 2

The cross-section and the volume of a plunge pool

A pool is a prism: every cross-section along its width is the same. The area of one end face, found by integration, times the width gives the volume of water it holds, and the curved floor lies below the water surface, so the integral is negative and the area is its modulus.

Not in G2 (G3 Additional Maths only): this uses a region below the $x$-axis (or left of the $y$-axis). The end of a plunge pool is shaped like the region between the water surface, the $x$-axis, and a curved floor with equation $y = \frac{x^{2}}{4} - 2 x$ for $0 \le x \le 8$, where $x$ and $y$ are measured in metres and the floor is below the water surface. The pool is $6$ metres wide, with the same end all the way across.

• (a)Find the area of the end of the pool, in square metres.
• (b)Hence find the volume of water the pool holds when it is full to the surface, in cubic metres.

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Colour keythe evaluated area or value

(a)

$\displaystyle \dfrac{0 + 8}{2} = 4$

The middle of the piece is a safe test value: it cannot be a root.

$\displaystyle \frac{\left(4\right)^{2}}{4} - 2 \times \left(4\right) = -4$

The curve's height at the test value: negative, so this piece lies below the $x$-axis.

$\displaystyle \int_{0}^{8} \left(\frac{x^{2}}{4} - 2 x\right)\,dx = \left[\frac{x^{3}}{12} - x^{2}\right]_{0}^{8}$

Integrate with the usual rules but leave out $c$: it cancels in a definite integral.

$\displaystyle = \left(\frac{\left(8\right)^{3}}{12} - \left(8\right)^{2}\right) - \left(\frac{\left(0\right)^{3}}{12} - \left(0\right)^{2}\right)$

Substitute the upper limit $8$, then the lower limit $0$, and subtract (upper minus lower).

$\displaystyle = \textcolor{#010106}{- \frac{64}{3}}$

Evaluate.

$\displaystyle \left|- \frac{64}{3}\right| = \textcolor{#010106}{\frac{64}{3}}$

An area is never negative: take the modulus of the negative integral.

(b)

Every cross-section along the width of the pool is the same, so the volume is the cross-section area times the width.

A prism: volume = (area of the end face) $\times$ (length).

$\displaystyle V\;= \frac{64}{3} \times 6$

Use the area from the first part.

$\displaystyle = \textcolor{#010106}{128}$

Multiply.

Answer

(a) $21.3$ m² (3 s.f.), exactly $\frac{64}{3}$

(b) $V = 128$ m³

$A$$B$$O$$\text{fence}$$\text{stream bank}$$\text{plan view}$

Maths In Real Life 3

A lawn between a fence and a stream

A park keeper has to order turf for a lawn with one straight edge and one curved edge. Turf is sold by the square metre, so the area has to be found before the order.

A lawn in a park lies between a straight fence and a curved stream bank, seen from above. Take the corner post $O$ as the origin, with $x$ and $y$ in metres. The stream bank is the curve $y = \frac{x^{2}}{2}$ and the fence is the line $y = 2 x + 6$. They meet at the points $A$ and $B$.

• (a)Find the coordinates of $A$ and $B$, and the area of the lawn.

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Colour keythe evaluated area or value

(a)

$\displaystyle \frac{x^{2}}{2} = 2 x + 6$

Where the line meets the curve the two $y$-values are equal.

$\displaystyle \frac{x^{2}}{2} - 2 x - 6 = 0$

Collect every term on one side.

$\displaystyle x = -2 \text{ or } x = 6$

Solve (factorise) to get the $x$-coordinate of each point.

$\displaystyle \dfrac{-2 + 6}{2} = 2$

Test value: the middle of the two intersections.

$\displaystyle \frac{\left(2\right)^{2}}{2} - \left(2 \times \left(2\right) + 6\right) = -8$

Curve height minus line height: negative, so the line is the upper boundary here.

$\displaystyle \int_{-2}^{6} \left(- \frac{x^{2}}{2} + 2 x + 6\right)\,dx = \left[- \frac{x^{3}}{6} + x^{2} + 6 x\right]_{-2}^{6}$

Integrate (upper $-$ lower) $= - \frac{x^{2}}{2} + 2 x + 6$, with no constant.

$\displaystyle = \left(- \frac{\left(6\right)^{3}}{6} + \left(6\right)^{2} + 6 \times \left(6\right)\right) - \left(- \frac{\left(-2\right)^{3}}{6} + \left(-2\right)^{2} + 6 \times \left(-2\right)\right)$

Substitute the upper limit $6$, then the lower limit $-2$, and subtract (upper minus lower).

$\displaystyle = \textcolor{#010106}{\frac{128}{3}}$

Evaluate.

Answer

(a) $(-2, 2)$, $(6, 18)$ (x and y in metres); $42.7$ m² (3 s.f.), exactly $\frac{128}{3}$

The symbol $\displaystyle\int_{a}^{b} f(x)\,dx$ is read “the integral of $f(x)$ from $a$ to $b$”: $a$ is the lower limit and $b$ is the upper limit. To evaluate it, find an antiderivative $F(x)$, substitute the upper limit, substitute the lower limit and subtract.

Roots and reciprocals are rewritten in index form first, exactly as in Chapter 14: $\dfrac{1}{x^{2}} = x^{-2}$ and $\sqrt{x} = x^{1/2}$. Then every term is integrated by adding $1$ to the power and dividing by the new power.

Not in G2 (G3 Additional Maths only): an integrand with $e^{ax}$ or a logarithm is G3 Additional Maths only. The limit $\ln n$ is chosen so that $e^{\ln n} = n$, and the answer comes out as a whole number.

A definite integral also obeys three rules. They let you work out new integrals from ones you are told, without knowing the function at all. The table shows each rule as a picture of areas.

When both limits are equal the integral is $0$, because there is no width. A constant multiplier comes outside the integral, and a constant term is integrated on its own: $\displaystyle\int_{a}^{b} k\,dx = k(b - a)$.

The same rules work when the given values are letters. Nothing changes except that the answers are expressions in $p$ and $q$ instead of numbers.

Not in G2 (G3 Additional Maths only): a fraction whose integral is a logarithm ($\dfrac{1}{x + 1}$ integrates to $\ln|x + 1|$) is G3 Additional Maths only.

A fraction with a linear bottom cannot be integrated as it stands. Divide the top by the bottom to rewrite it as a polynomial plus a fraction with a constant top, $\dfrac{N}{D} = Q + \dfrac{R}{D}$; the polynomial is integrated by the usual rule and the fraction gives a logarithm.

Cut the region under a curve into very thin vertical strips. Each strip is almost a rectangle of height $y$ and tiny width $dx$, so its area is about $y\,dx$. Adding all the strips from $x = a$ to $x = b$ is exactly what the integral does.

The limits must not include a value where the curve is undefined. The curve $y = \dfrac{k}{x^{2}}$ does not exist at $x = 0$, so an area under it must start and stop on the same side of $0$.

Not in G2 (G3 Additional Maths only): the exponential $e^{ax}$ and the logarithm from $\dfrac{1}{x}$ are G3 Additional Maths only. These two Examples give exact answers, because the limit is $\ln n$.

When a question says sketch, the picture is part of the answer. Mark where the curve meets the axes, check that it lies above the $x$-axis between the limits, and shade the region. The sketch also protects you from integrating over a stretch where the curve dips below the axis.