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ISC 2027
All chaptersMaths · Unit 3

Applications of Integrals

4 articles16 formulas20 ways the board asks it
MATArea under a Curve

Area under a curve (between curve and an axis)

This subtopic uses the definite integral to compute the area of a region bounded by a single curve, one coordinate axis and a pair of bounding lines (ordinates or abscissae). The core idea is that ∫aby dx\int_a^b y\,dx measures the signed area of vertical strips between the curve y=f(x)y=f(x) and the x-axis, while ∫cdx dy\int_c^d x\,dy measures horizontal strips against the y-axis.

It is the foundational skill on which every other area problem in the chapter is built.

Area against the x-axis
A=∫aby dx=∫abf(x) dxA = \int_{a}^{b} y\,dx = \int_{a}^{b} f(x)\,dx
y=f(x)y=f(x) lies above the x-axis on [a,b][a,b], with a<ba<b; AA is the area between the curve and the x-axis from x=ax=a to x=bx=b.
Area against the y-axis
A=∫cdx dy=∫cdg(y) dyA = \int_{c}^{d} x\,dy = \int_{c}^{d} g(y)\,dy
x=g(y)x=g(y) is the curve expressed as a function of yy, lying to the right of the y-axis on [c,d][c,d], with c<dc<d.
Area when the curve dips below the x-axis
A=∣∫abf(x) dx∣A = \left| \int_{a}^{b} f(x)\,dx \right|
used when f(x)≤0f(x)\le 0 on [a,b][a,b], so the raw integral is negative; the magnitude gives the physical area.
Splitting at a sign change
A=∣∫acf(x) dx∣+∣∫cbf(x) dx∣A = \left| \int_{a}^{c} f(x)\,dx \right| + \left| \int_{c}^{b} f(x)\,dx \right|
x=cx=c is the point in (a,b)(a,b) where f(x)f(x) changes sign (crosses the x-axis); each piece is integrated separately and the magnitudes are added.
  • Decide the strip direction first: use ∫y dx\int y\,dx for regions bounded by the x-axis and ordinates, ∫x dy\int x\,dy for regions bounded by the y-axis and horizontal lines.
  • Always sketch the curve and shade the region; the sketch tells you the limits and whether the area lies above or below the axis.
  • Area is intrinsically positive: if the region is below the x-axis the integral is negative, so report ∣A∣|A|.
  • If the curve crosses the axis inside [a,b][a,b], split the integral at the crossing point and add the absolute values of the parts; do NOT integrate straight through.
  • For ∫x dy\int x\,dy rewrite the curve as x=g(y)x=g(y) (e.g. y2=x⇒x=y2y^2=x \Rightarrow x=y^2) and use the y-limits.
  • The bounding lines x=a, x=bx=a,\,x=b (or y=c, y=dy=c,\,y=d) supply the limits of integration directly.
  • Standard results worth recalling: ∫0πsin⁡x dx=2\int_0^\pi \sin x\,dx = 2 and ∫02x2 dx=83\int_0^2 x^2\,dx = \dfrac{8}{3}.
  • Units of area follow from the context (e.g. square units); leave a pure number if no units are given.
Where the marks go
  • Computing ∫02πsin⁡x dx=0\int_0^{2\pi}\sin x\,dx = 0 and reporting zero area, instead of splitting at x=πx=\pi to get total area 44.
  • Forgetting to take the absolute value when the region lies below the x-axis (e.g. y=cos⁡xy=\cos x on [π2,π]\left[\dfrac{\pi}{2},\pi\right]), giving a negative 'area'.
  • Using ∫y dx\int y\,dx when the region is bounded by the y-axis and lines y=c, y=dy=c,\,y=d, where ∫x dy\int x\,dy is required.
  • Swapping or mislabelling the limits so that a>ba>b, which flips the sign of the answer.
How the board asks it
  • Numericalvertical strips against the x-axis using ∫aby dx\int_a^b y\,dx
    Find the area of the region bounded by the curve y=x2+2y = x^2 + 2, the x-axis and the ordinates x=1x = 1 and x=3x = 3.
  • Diagram / graphthe sketch fixes the limits and tells you if the region is above or below the axis
    Draw a rough sketch of the curve y=sin⁡xy = \sin x for 0≤x≤2π0 \le x \le 2\pi and hence find the total area enclosed between the curve and the x-axis.
  • Numericalsplitting at the crossing point and adding the absolute values of the parts
    Find the area bounded by the curve y=x3y = x^3, the x-axis and the lines x=−2x = -2 and x=1x = 1, taking into account the portion lying below the x-axis.
  • Numericalhorizontal strips against the y-axis using ∫cdx dy\int_c^d x\,dy with x=g(y)x = g(y)
    Find the area of the region bounded by the parabola y2=4xy^2 = 4x, the y-axis and the lines y=1y = 1 and y=3y = 3.
  • Give reasonsa below-axis integral is negative while area is intrinsically positive
    Explain, with the help of a sketch, why ∫02πsin⁡x dx=0\int_0^{2\pi} \sin x\,dx = 0 does not give the area enclosed between y=sin⁡xy = \sin x and the x-axis on [0,2π][0, 2\pi], and state the correct area.

Practise this topic

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.