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How do definite integrals measure the area enclosed between two curves, even when they cross?

Find the area between curves using definite integration, accounting for intersection points and which curve is on top

WACE Specialist Unit 4 areas between curves: the top-minus-bottom integral, finding intersection points as limits, splitting the region where curves cross, and integrating with respect to y when convenient, with a worked example.

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  1. What this dot point is asking
  2. Top minus bottom
  3. Finding the limits
  4. Why top minus bottom always works
  5. When the curves cross inside
  6. Integrating with respect to y
  7. Choosing the right strip orientation
  8. A note on signed area versus enclosed area

What this dot point is asking

SCSA wants you to set up and evaluate the area between two curves, locate the boundaries by solving for intersections, and handle regions where the upper and lower curves change.

Top minus bottom

The integrand is the vertical gap between the curves at each xx. Because we subtract the lower curve, the result is positive and there is no need to worry about whether either curve dips below the xx-axis: the difference handles that automatically.

Finding the limits

The limits aa and bb are usually the xx-coordinates of the intersection points, found by solving f(x)=g(x)f(x) = g(x). Sketch the region to confirm which curve is on top, since that determines the order of subtraction.

Why top minus bottom always works

The reason the formula needs no absolute-value signs is geometric. At a fixed xx, the height of the thin strip between the curves is the upper yy-value minus the lower yy-value, which is f(x)g(x)f(x) - g(x) whenever ff is on top. Summing these strip areas as the width dx0dx \to 0 is exactly the definite integral. Because f(x)g(x)0f(x) - g(x) \ge 0 throughout the interval, the integral is positive automatically, even where both curves are below the xx-axis, because shifting both curves up by the same amount leaves their difference unchanged. This is why you should never integrate ff and gg separately and subtract the signed areas: subtract the functions first, then integrate once.

When the curves cross inside

Integrating with respect to y

When the region is bounded more naturally on the left and right, it can be cleaner to write xx as a function of yy and integrate cd[xrightxleft]dy\int_c^d [x_{\text{right}} - x_{\text{left}}]\,dy. Choose the variable that makes the bounds simplest. A region trapped between a sideways parabola x=y2x = y^2 and a line, for instance, has a single clean description in terms of yy but would need splitting in terms of xx, so the choice of variable can turn two integrals into one.

Choosing the right strip orientation

Deciding whether to integrate in xx or in yy is a judgement worth making before any algebra. Use vertical strips (integrate in xx) when the region has a clear top curve and bottom curve over a single xx-interval. Use horizontal strips (integrate in yy) when the region has a clear right boundary and left boundary over a single yy-interval, which is typical when the curves are given as xx in terms of yy or open sideways. If neither orientation avoids a split, pick whichever produces the simpler integrand. A quick sketch with one sample strip drawn in shows immediately which curve is the upper or right boundary and which is the lower or left boundary.

A note on signed area versus enclosed area

The phrase area between curves always means a positive, enclosed area, so the top-minus-bottom rule with a split at every crossing is the safe method. This is different from a single definite integral abf(x)dx\int_a^b f(x)\,dx, which is a signed area that can be negative where ff lies below the axis. SCSA questions are explicit about which they want: an enclosed area is always positive, whereas a net signed value may be requested in a motion or accumulation context. Reading the command word carefully decides whether to split for positivity or to leave the integral signed.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20227 marksCalculator-assumed. (a) Find the points of intersection of y=x3xy = x^3 - x and y=xy = x. (b) Hence find the total area enclosed between the two curves.
Show worked answer →

A curves-that-cross area, requiring a split.

(a) Solve x3x=xx^3 - x = x, so x32x=0x^3 - 2x = 0, x(x22)=0x(x^2 - 2) = 0, giving x=0x = 0, x=2x = \sqrt{2} and x=2x = -\sqrt{2}.

(b) Between 2-\sqrt{2} and 00 the cubic is above the line; between 00 and 2\sqrt{2} the line is above the cubic. By the symmetry of these odd functions the two pieces have equal area. On [0,2][0, \sqrt{2}] the integrand is x(x3x)=2xx3x - (x^3 - x) = 2x - x^3, so 02(2xx3)dx=[x2x44]02=244=1\displaystyle\int_0^{\sqrt{2}} (2x - x^3)\,dx = \left[x^2 - \dfrac{x^4}{4}\right]_0^{\sqrt{2}} = 2 - \dfrac{4}{4} = 1. The total area is 2×1=22 \times 1 = 2 square units.

Markers reward the three intersection points, recognising the swap of upper and lower curve at x=0x = 0, the split (or symmetry argument), and the total area 22.

WACE 20245 marksCalculator-free. The region is bounded by the curve x=y2x = y^2 and the line x=y+2x = y + 2. Find its area by integrating with respect to yy.
Show worked answer →

An integrate-with-respect-to-yy question.

Intersections: y2=y+2y^2 = y + 2, so y2y2=0y^2 - y - 2 = 0, (y2)(y+1)=0(y - 2)(y + 1) = 0, giving y=2y = 2 and y=1y = -1. For yy between 1-1 and 22 the line x=y+2x = y + 2 is to the right of the parabola x=y2x = y^2.

Area =12[(y+2)y2]dy=[y22+2yy33]12= \displaystyle\int_{-1}^{2} \big[(y + 2) - y^2\big]\,dy = \left[\dfrac{y^2}{2} + 2y - \dfrac{y^3}{3}\right]_{-1}^{2}. At y=2y = 2: 2+483=1032 + 4 - \dfrac{8}{3} = \dfrac{10}{3}. At y=1y = -1: 122+13=76\dfrac{1}{2} - 2 + \dfrac{1}{3} = -\dfrac{7}{6}. Difference: 103+76=276=92\dfrac{10}{3} + \dfrac{7}{6} = \dfrac{27}{6} = \dfrac{9}{2}.

Markers reward the limits in yy, right-curve minus left-curve, and the area 92\dfrac{9}{2} square units.

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