Calculus · Paper 1
Integration — Part 3
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Section 1 of 14
Definite integrals — the concept
In Part 1 we saw that area under a curve = anti-derivative at the right end − anti-derivative at the left end. That's so important it gets its own notation.
Must learn — definite integral notation
$\displaystyle\int_{a}^{b} f(x)\,dx = \Big[\,F(x)\,\Big]_{a}^{b} = F(b) - F(a)$
$F(x)$ is any anti-derivative of $f(x)$ · $a$ is the lower limit, $b$ the upper.
No $+c$ needed. The constants would cancel: $(F(b) + c) - (F(a) + c) = F(b) - F(a)$.
No $+c$ needed. The constants would cancel: $(F(b) + c) - (F(a) + c) = F(b) - F(a)$.
(i) $\displaystyle\int_{0}^{1} x\,dx$
$\displaystyle\int_{0}^{1} x\,dx = \Big[\,\dfrac{x^{2}}{2}\,\Big]_{0}^{1}$
$= \dfrac{1^{2}}{2} - \dfrac{0^{2}}{2}$
$= \dfrac{1}{2} - 0 = \dfrac{1}{2}$
Check: triangle of base $1$ and height $1$ has area $\tfrac{1}{2}(1)(1) = \dfrac{1}{2}$. ✓
(ii) $\displaystyle\int_{0}^{3} 5\,dx$
$\displaystyle\int_{0}^{3} 5\,dx = \Big[\,5x\,\Big]_{0}^{3} = 5(3) - 5(0) = 15$
Check: rectangle of width $3$ and height $5$ has area $15$. ✓
(iii) $\displaystyle\int_{0}^{1} e^{x}\,dx$
$\displaystyle\int_{0}^{1} e^{x}\,dx = \Big[\,e^{x}\,\Big]_{0}^{1} = e^{1} - e^{0} = e - 1$
No nice formula for that area — but the integral gives it exactly.
You try
Evaluate $\displaystyle\int_{1}^{3} (2x + 1)\,dx$.
Anti-derivative is $x^{2} + x$. Evaluate at $3$, then at $1$, subtract.
$\Big[\,x^{2} + x\,\Big]_{1}^{3}$
$= (9 + 3) - (1 + 1)$
$= 12 - 2$
$= 10$
$10$
You try — one more
Evaluate $\displaystyle\int_{0}^{2} e^{2x}\,dx$.
Anti-derivative of $e^{2x}$ is $\dfrac{1}{2}\,e^{2x}$.
$\Big[\,\dfrac{1}{2}\,e^{2x}\,\Big]_{0}^{2}$
$= \dfrac{1}{2}\,e^{4} - \dfrac{1}{2}\,e^{0}$
$= \dfrac{1}{2}(e^{4} - 1)$
$\dfrac{1}{2}(e^{4} - 1)$
Section 2 of 14
Area between a line and the $x$-axis
From here on, the question usually says: "Find the area of the region bounded by...". Recipe doesn't change — integrate, evaluate, subtract. Just keep track of which curve and which limits.
Worked — area between $y = 2x - 1$ and the $x$-axis from $x = 2$ to $x = 5$
$A = \displaystyle\int_{2}^{5} (2x - 1)\,dx$
$= \Big[\,x^{2} - x\,\Big]_{2}^{5}$
$= (25 - 5) - (4 - 2)$
$= 20 - 2$
$A = 18$ sq units
Check. The region is a trapezium with parallel sides $3$ (height at $x=2$) and $9$ (height at $x=5$), separated by horizontal distance $3$.
$A = \left(\dfrac{3 + 9}{2}\right)(3) = 6 \times 3 = 18$ ✓
You try
Find the area of the region bounded by $y = 3x + 2$, the $x$-axis, and the lines $x = 1$ and $x = 4$.
$A = \displaystyle\int_{1}^{4} (3x + 2)\,dx$. Anti-derivative: $\dfrac{3x^{2}}{2} + 2x$.
$A = \displaystyle\int_{1}^{4} (3x + 2)\,dx$
$= \Big[\,\dfrac{3x^{2}}{2} + 2x\,\Big]_{1}^{4}$
$= (24 + 8) - (\dfrac{3}{2} + 2)$
$= 32 - \dfrac{7}{2}$
$= \dfrac{57}{2} = 28.5$ sq units
$28.5$ sq units
Section 3 of 14
When the curve dips below — split and mod
Area is always positive. The integral can be negative. If the curve goes below the $x$-axis, the integral over that bit comes out negative — because we're "subtracting" heights below.
The trick: split the integral at the crossing point, and take the modulus (absolute value) of each piece. Then add them.
Worked — area between $y = x - 1$ and the $x$-axis from $x = 0$ to $x = 2$
The curve crosses $y = 0$ when $x = 1$. That's the split point.
Region $A_{1}$: $\displaystyle\int_{0}^{1} (x - 1)\,dx = \Big[\,\dfrac{x^{2}}{2} - x\,\Big]_{0}^{1}$
$= \left(\dfrac{1}{2} - 1\right) - (0)$
$= -\dfrac{1}{2}$ (negative — region is below the axis)
$|A_{1}| = \dfrac{1}{2}$
Region $A_{2}$: $\displaystyle\int_{1}^{2} (x - 1)\,dx = \Big[\,\dfrac{x^{2}}{2} - x\,\Big]_{1}^{2}$
$= (2 - 2) - \left(\dfrac{1}{2} - 1\right)$
$= 0 - \left(-\dfrac{1}{2}\right) = \dfrac{1}{2}$
Total area $= |A_{1}| + A_{2} = \dfrac{1}{2} + \dfrac{1}{2} = 1$ sq unit
The rule — when the curve crosses the axis
(1) Find where the curve crosses the $x$-axis. That's your split point.
(2) Integrate each piece separately.
(3) Take the modulus of any negative result.
(4) Add them.
(2) Integrate each piece separately.
(3) Take the modulus of any negative result.
(4) Add them.
Famous mistake. Just integrating $\displaystyle\int_{0}^{2}(x-1)\,dx$ from $0$ to $2$ in one go would give $0$ — the two halves cancel. The integral is right, the area is wrong.
You try
Find the area enclosed by $y = x - 3$, the $x$-axis, and the lines $x = 1$ and $x = 5$.
The line crosses $y=0$ at $x=3$. Split into $[1,3]$ (below) and $[3,5]$ (above).
$A_{1}$ on $[1,3]$: $\Big[\,\dfrac{x^{2}}{2} - 3x\,\Big]_{1}^{3} = \left(\dfrac{9}{2} - 9\right) - \left(\dfrac{1}{2} - 3\right) = -\dfrac{9}{2} - (-\dfrac{5}{2}) = -2$
$|A_{1}| = 2$
$A_{2}$ on $[3,5]$: $\Big[\,\dfrac{x^{2}}{2} - 3x\,\Big]_{3}^{5} = \left(\dfrac{25}{2} - 15\right) - \left(\dfrac{9}{2} - 9\right) = -\dfrac{5}{2} - (-\dfrac{9}{2}) = 2$
Total $= 2 + 2 = 4$ sq units
$A = 4$ sq units
$4$ sq units
Section 4 of 14
Area between a quadratic and the $x$-axis
Same recipe — but the curve does more interesting things. Three patterns to recognise:
(i) Quadratic entirely below the $x$-axis — $y = x^{2} - 3x$ between its roots
Find the roots: $x^{2} - 3x = 0 \;\Rightarrow\; x(x-3) = 0 \;\Rightarrow\; x = 0$ or $x = 3$.
Between $0$ and $3$ the curve sits below the axis, so the integral will come out negative.
$\displaystyle\int_{0}^{3} (x^{2} - 3x)\,dx = \Big[\,\dfrac{x^{3}}{3} - \dfrac{3x^{2}}{2}\,\Big]_{0}^{3}$
$= \left(\dfrac{27}{3} - \dfrac{27}{2}\right) - (0)$
$= 9 - \dfrac{27}{2} = -\dfrac{9}{2}$
$A = \left|-\dfrac{9}{2}\right| = \dfrac{9}{2}$ or $4.5$ sq units
(ii) Quadratic entirely above the axis — $y = 4 - x^{2}$ between its roots
Roots: $4 - x^{2} = 0 \;\Rightarrow\; x = \pm 2$. Curve sits above the axis between them.
$A = \displaystyle\int_{-2}^{2} (4 - x^{2})\,dx = \Big[\,4x - \dfrac{x^{3}}{3}\,\Big]_{-2}^{2}$
$= \left(8 - \dfrac{8}{3}\right) - \left(-8 + \dfrac{8}{3}\right)$
$= 8 - \dfrac{8}{3} + 8 - \dfrac{8}{3} = 16 - \dfrac{16}{3}$
$A = \dfrac{48 - 16}{3} = \dfrac{32}{3}$ sq units
(iii) Quadratic between two given $x$-values — $y = x^{2} + 4$ from $x = 1$ to $x = 3$
No need to find roots — they didn't ask. The limits $x=1$ and $x=3$ are the limits.
$A = \displaystyle\int_{1}^{3} (x^{2} + 4)\,dx = \Big[\,\dfrac{x^{3}}{3} + 4x\,\Big]_{1}^{3}$
$= \left(9 + 12\right) - \left(\dfrac{1}{3} + 4\right)$
$= 21 - \dfrac{13}{3}$
$A = \dfrac{63 - 13}{3} = \dfrac{50}{3}$ sq units
You try
Find the area enclosed by $y = x^{2} - 4x$ and the $x$-axis.
Roots: $x(x-4)=0 \Rightarrow x = 0, 4$. Curve sits below — integral comes out negative, take mod.
$\displaystyle\int_{0}^{4} (x^{2} - 4x)\,dx = \Big[\,\dfrac{x^{3}}{3} - 2x^{2}\,\Big]_{0}^{4}$
$= \left(\dfrac{64}{3} - 32\right) - 0 = \dfrac{64 - 96}{3} = -\dfrac{32}{3}$
$A = \dfrac{32}{3}$ sq units
$\dfrac{32}{3}$ sq units
Section 5 of 14
Quadratic that crosses the axis — split & mod
If the given limits straddle a root, part of the region is below the axis and part is above. Same idea as Section 3 — split at the root, mod each piece.
Worked — area between $y = x^{2} - 4$ and the $x$-axis from $x = 1$ to $x = 3$
$x^{2} - 4 = 0$ at $x = \pm 2$. The relevant root in our range $[1, 3]$ is $x = 2$ — that's the split point.
$A_{1}$ on $[1, 2]$: $\displaystyle\int_{1}^{2} (x^{2} - 4)\,dx = \Big[\,\dfrac{x^{3}}{3} - 4x\,\Big]_{1}^{2}$
$= \left(\dfrac{8}{3} - 8\right) - \left(\dfrac{1}{3} - 4\right)$
$= \dfrac{8 - 24}{3} - \dfrac{1 - 12}{3} = -\dfrac{16}{3} + \dfrac{11}{3} = -\dfrac{5}{3}$
$|A_{1}| = \dfrac{5}{3}$
$A_{2}$ on $[2, 3]$: $\displaystyle\int_{2}^{3} (x^{2} - 4)\,dx = \Big[\,\dfrac{x^{3}}{3} - 4x\,\Big]_{2}^{3}$
$= \left(9 - 12\right) - \left(\dfrac{8}{3} - 8\right) = -3 - \left(-\dfrac{16}{3}\right) = -3 + \dfrac{16}{3} = \dfrac{7}{3}$
Total area $= \dfrac{5}{3} + \dfrac{7}{3} = \dfrac{12}{3} = 4$ sq units
You try
Find the area enclosed between $y = x^{2} - 1$, the $x$-axis, and the lines $x = 0$ and $x = 2$.
Roots at $x = \pm 1$. Relevant split point in $[0,2]$ is $x = 1$.
$A_{1}$ on $[0,1]$: $\Big[\,\dfrac{x^{3}}{3} - x\,\Big]_{0}^{1} = \left(\dfrac{1}{3} - 1\right) - 0 = -\dfrac{2}{3}$ → $|A_{1}| = \dfrac{2}{3}$
$A_{2}$ on $[1,2]$: $\Big[\,\dfrac{x^{3}}{3} - x\,\Big]_{1}^{2} = \left(\dfrac{8}{3} - 2\right) - \left(\dfrac{1}{3} - 1\right) = \dfrac{2}{3} - (-\dfrac{2}{3}) = \dfrac{4}{3}$
Total $= \dfrac{2}{3} + \dfrac{4}{3} = 2$
$A = 2$ sq units
$2$ sq units
Section 6 of 14
Cubic with sign changes
Same idea, more roots. A cubic can cross the $x$-axis up to three times, giving up to three sign-change pieces.
Worked — area enclosed by $y = 12x^{3} - 48x^{2} + 36x$ and the $x$-axis
Step 1 — find the roots.
$12x^{3} - 48x^{2} + 36x = 0$
$12x(x^{2} - 4x + 3) = 0$
$12x(x - 1)(x - 3) = 0$
Roots: $x = 0, \;\; x = 1, \;\; x = 3$
Between $0$ and $1$, the curve is above the axis. Between $1$ and $3$, below.
Step 2 — anti-differentiate once.
$\displaystyle\int (12x^{3} - 48x^{2} + 36x)\,dx = 3x^{4} - 16x^{3} + 18x^{2}$
Step 3 — evaluate each piece.
$A_{1}$ on $[0, 1]$:
$\Big[\,3x^{4} - 16x^{3} + 18x^{2}\,\Big]_{0}^{1} = (3 - 16 + 18) - 0 = 5$
$A_{2}$ on $[1, 3]$:
$\Big[\,3x^{4} - 16x^{3} + 18x^{2}\,\Big]_{1}^{3}$
$= (3(81) - 16(27) + 18(9)) - (3 - 16 + 18)$
$= (243 - 432 + 162) - 5$
$= -27 - 5 = -32$
$|A_{2}| = 32$
Total area $= 5 + 32 = 37$ sq units
You try
Find the total area enclosed by $y = x^{3} - x$ and the $x$-axis.
$x^{3} - x = x(x-1)(x+1)$. Roots: $-1, 0, 1$. Two regions: $[-1, 0]$ and $[0, 1]$. Symmetry helps.
Anti-derivative: $\dfrac{x^{4}}{4} - \dfrac{x^{2}}{2}$
$A_{1}$ on $[-1, 0]$: $\Big[\,\dfrac{x^{4}}{4} - \dfrac{x^{2}}{2}\,\Big]_{-1}^{0} = 0 - \left(\dfrac{1}{4} - \dfrac{1}{2}\right) = \dfrac{1}{4}$
$A_{2}$ on $[0, 1]$: $\Big[\,\dfrac{x^{4}}{4} - \dfrac{x^{2}}{2}\,\Big]_{0}^{1} = \left(\dfrac{1}{4} - \dfrac{1}{2}\right) - 0 = -\dfrac{1}{4}$ → $|A_{2}| = \dfrac{1}{4}$
Total $= \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2}$
$A = \dfrac{1}{2}$ sq unit
$\dfrac{1}{2}$ sq unit
Section 7 of 14
Area between two curves — top minus bottom
When the region is bounded above by one curve and below by another, the area is the integral of the difference. Top curve minus bottom curve.
Must learn
$A = \displaystyle\int_{a}^{b} \big(f_{\text{top}}(x) - f_{\text{bottom}}(x)\big)\,dx$
$a$ and $b$ are the $x$-values where the curves meet — solve $f_{\text{top}} = f_{\text{bottom}}$ to find them.
(i) Area between $y = x^{2}$ and $y = x$
Find the intersections. Set the curves equal:
$x^{2} = x$
$x^{2} - x = 0$
$x(x - 1) = 0 \;\;\Rightarrow\;\; x = 0, \;\; x = 1$
Between $0$ and $1$, which is on top? At $x = \tfrac{1}{2}$: $y = x$ gives $\tfrac{1}{2}$, $y = x^{2}$ gives $\tfrac{1}{4}$. So $y = x$ is on top.
$A = \displaystyle\int_{0}^{1} (x - x^{2})\,dx = \Big[\,\dfrac{x^{2}}{2} - \dfrac{x^{3}}{3}\,\Big]_{0}^{1}$
$= \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{3 - 2}{6} = \dfrac{1}{6}$
$A = \dfrac{1}{6}$ sq unit
(ii) Two intersecting parabolas — $f(x) = 12 - 3x^{2}$ and $g(x) = 9x^{2}$
Find the area enclosed between the two curves, then show that the line $y = 9$ divides this area in the ratio $1:3$.
Find the intersections. $f = g$ gives:
$12 - 3x^{2} = 9x^{2} \;\Rightarrow\; 12 = 12x^{2} \;\Rightarrow\; x = \pm 1$
At $x = \pm 1$, $y = 9(1)^{2} = 9$. So the curves meet on the line $y = 9$.
Total area — $f$ on top (concave-down), $g$ on bottom (concave-up):
$A_{\text{total}} = \displaystyle\int_{-1}^{1} (f - g)\,dx = \displaystyle\int_{-1}^{1} \big[(12 - 3x^{2}) - 9x^{2}\big]\,dx$
$= \displaystyle\int_{-1}^{1} (12 - 12x^{2})\,dx = \Big[\,12x - 4x^{3}\,\Big]_{-1}^{1}$
$= (12 - 4) - (-12 + 4) = 8 - (-8) = 16$ sq units
Top piece $A_{1}$ — between $f$ and the line $y = 9$:
$A_{1} = \displaystyle\int_{-1}^{1} (f - 9)\,dx = \displaystyle\int_{-1}^{1} (3 - 3x^{2})\,dx$
$= \Big[\,3x - x^{3}\,\Big]_{-1}^{1} = (3 - 1) - (-3 + 1) = 2 - (-2) = 4$ sq units
Bottom piece $A_{2}$ — between $y = 9$ and $g$:
$A_{2} = \displaystyle\int_{-1}^{1} (9 - g)\,dx = \displaystyle\int_{-1}^{1} (9 - 9x^{2})\,dx$
$= \Big[\,9x - 3x^{3}\,\Big]_{-1}^{1} = (9 - 3) - (-9 + 3) = 6 - (-6) = 12$ sq units
Check: $A_{1} + A_{2} = 4 + 12 = 16 = A_{\text{total}}$ ✓
Ratio: $A_{1} : A_{2} = 4 : 12 = 1 : 3$ ✓
The line $y = 9$ divides the enclosed region in the ratio $1:3$ — the bottom piece is three times the top.
You try
Find the area enclosed between $y = x^{2}$ and $y = 2x$.
Intersections: $x^{2} = 2x \Rightarrow x = 0, 2$. Between them, $2x > x^{2}$, so line is on top.
$A = \displaystyle\int_{0}^{2} (2x - x^{2})\,dx = \Big[\,x^{2} - \dfrac{x^{3}}{3}\,\Big]_{0}^{2}$
$= 4 - \dfrac{8}{3} = \dfrac{12 - 8}{3} = \dfrac{4}{3}$
$A = \dfrac{4}{3}$ sq units
$\dfrac{4}{3}$ sq units
Section 8 of 14
Tangent line + curve
Sometimes the bounding line is a tangent to the curve — it touches the curve at exactly one point. Same technique: top minus bottom, integrate, evaluate.
(i) The line $2x - y - 10 = 0$ is a tangent to $y = x^{2} - 9$. Find the area between the curve, the tangent, and the $x$-axis.
Strategy. Tangent touches at $(1, -8)$, the curve crosses the $x$-axis at $x = 3$, and the tangent crosses the $x$-axis at $x = 5$. The required region is bounded above by the $x$-axis, on the left by the curve from $x = 3$ down to its lowest part, then back up along the tangent to $(5, 0)$.
Approach: find the area below the $x$-axis bounded by the curve from $1$ to $3$, plus the triangle bounded by the tangent and the $x$-axis from $1$ to $5$.
This kind of question is geometric — always sketch first, then split the region.
(ii) Tangent at a local maximum — $f(x) = x^{3} - 3x^{2} + 5$
At a local max, the tangent is horizontal — a constant. That makes the calculation simple: it's $\int$(constant $-$ cubic) on the relevant range.
Find the local max. $f'(x) = 3x^{2} - 6x = 3x(x - 2) = 0$. So $x = 0$ or $x = 2$. The local max is at $x = 0$ (where $f''(0) = -6 < 0$). At $x = 0$, $f(0) = 5$. So the horizontal tangent is $y = 5$.
The line $y = 5$ also crosses the curve where $x^{3} - 3x^{2} + 5 = 5$, i.e. $x^{3} - 3x^{2} = 0$, so $x^{2}(x - 3) = 0$ → $x = 0$ (touch) and $x = 3$ (cross).
$A = \displaystyle\int_{0}^{3} \big[5 - (x^{3} - 3x^{2} + 5)\big]\,dx$
$= \displaystyle\int_{0}^{3} (3x^{2} - x^{3})\,dx = \Big[\,x^{3} - \dfrac{x^{4}}{4}\,\Big]_{0}^{3}$
$= \left(27 - \dfrac{81}{4}\right) - 0 = \dfrac{108 - 81}{4} = \dfrac{27}{4}$
$A = \dfrac{27}{4}$ sq units
You try
$f(x) = -x^{3} + 6x^{2} - 9x + 6$ has a local maximum. Find the area enclosed between the horizontal tangent at that max and the curve.
$f'(x) = -3x^{2} + 12x - 9 = -3(x-1)(x-3)$. Local max at $x=3$ ($f''(3)<0$): $f(3) = -27 + 54 - 27 + 6 = 6$. So tangent is $y = 6$. Other intersection: $f(x) = 6 \Rightarrow x = 0$ or $x = 3$.
Local max at $x = 3$: $f(3) = 6$. Horizontal tangent $y = 6$.
$f(x) = 6 \Rightarrow -x^{3} + 6x^{2} - 9x = 0 \Rightarrow -x(x-3)^{2} = 0$, so $x = 0$ or $x = 3$.
$A = \displaystyle\int_{0}^{3} \big[6 - (-x^{3} + 6x^{2} - 9x + 6)\big]\,dx = \displaystyle\int_{0}^{3} (x^{3} - 6x^{2} + 9x)\,dx$
$= \Big[\,\dfrac{x^{4}}{4} - 2x^{3} + \dfrac{9x^{2}}{2}\,\Big]_{0}^{3} = \dfrac{81}{4} - 54 + \dfrac{81}{2} = \dfrac{81 - 216 + 162}{4} = \dfrac{27}{4}$
$A = \dfrac{27}{4}$ sq units
$\dfrac{27}{4}$ sq units
Section 9 of 14
Curve + line — the teardrop and similar regions
Two intersecting curves (or a curve and a line) enclose a leaf- or teardrop-shaped region. Standard method: find the intersection points, take top minus bottom.
Worked — area enclosed by $y = 4 - x^{2}$ and the line $2x + y - 1 = 0$
Re-arrange the line. $2x + y - 1 = 0 \;\Rightarrow\; y = 1 - 2x$.
Find the intersections. $4 - x^{2} = 1 - 2x \;\Rightarrow\; x^{2} - 2x - 3 = 0 \;\Rightarrow\; (x-3)(x+1) = 0$. So $x = -1$ or $x = 3$.
Which is on top? At $x = 0$: parabola gives $4$, line gives $1$. Parabola is on top.
$A = \displaystyle\int_{-1}^{3} \big[(4 - x^{2}) - (1 - 2x)\big]\,dx$
$= \displaystyle\int_{-1}^{3} (3 + 2x - x^{2})\,dx$
$= \Big[\,3x + x^{2} - \dfrac{x^{3}}{3}\,\Big]_{-1}^{3}$
$= \left(9 + 9 - 9\right) - \left(-3 + 1 + \dfrac{1}{3}\right)$
$= 9 - \left(-\dfrac{5}{3}\right) = 9 + \dfrac{5}{3} = \dfrac{32}{3}$
$A = \dfrac{32}{3}$ sq units
You try
Find the area enclosed between $y = 6 - x^{2}$ and $y = x$.
Intersections: $6 - x^{2} = x \Rightarrow x^{2} + x - 6 = 0 \Rightarrow (x+3)(x-2) = 0$, so $x = -3, 2$.
At $x = 0$: parabola gives $6$, line gives $0$. Parabola on top.
$A = \displaystyle\int_{-3}^{2} \big[(6 - x^{2}) - x\big]\,dx = \displaystyle\int_{-3}^{2} (6 - x - x^{2})\,dx$
$= \Big[\,6x - \dfrac{x^{2}}{2} - \dfrac{x^{3}}{3}\,\Big]_{-3}^{2}$
$= \left(12 - 2 - \dfrac{8}{3}\right) - \left(-18 - \dfrac{9}{2} + 9\right) = \dfrac{22}{3} - \left(-\dfrac{27}{2}\right) = \dfrac{22}{3} + \dfrac{27}{2}$
$= \dfrac{44 + 81}{6} = \dfrac{125}{6}$
$A = \dfrac{125}{6}$ sq units
$\dfrac{125}{6}$ sq units
Section 10 of 14
Integration with respect to $y$
Sometimes the natural variable is $y$, not $x$ — for example when the region is bounded by the $y$-axis and horizontal lines. The formula flips: integrate $x$ as a function of $y$, with respect to $y$.
Must learn
$A = \displaystyle\int_{a}^{b} x \, dy$ (when integrating along the $y$-axis)
Rearrange the equation so $x$ is the subject. The limits are $y$-values, not $x$-values.
(i) Direct $y$-integration — $\displaystyle\int_{0}^{2} (3y + 1)\,dy$
$\Big[\,\dfrac{3y^{2}}{2} + y\,\Big]_{0}^{2} = (6 + 2) - 0 = 8$
(ii) The area between $y = x - 2$ and the $y$-axis from $y = 0$ to $y = 3$ — two ways
Method 1 — split with respect to $x$.
Trapezium with vertices $(0, 0), (2, 0), (5, 3), (0, 3)$. Parallel sides have lengths $2$ and $5$, height $3$.
$A = \left(\dfrac{2 + 5}{2}\right)(3) = \dfrac{21}{2}$ sq units
Method 2 — integrate with respect to $y$.
Rearrange the line: $y = x - 2 \;\Rightarrow\; x = y + 2$. That's $x$ as a function of $y$.
$A = \displaystyle\int_{0}^{3} (y + 2)\,dy = \Big[\,\dfrac{y^{2}}{2} + 2y\,\Big]_{0}^{3}$
$= \left(\dfrac{9}{2} + 6\right) - 0 = \dfrac{21}{2}$ sq units ✓
Both methods give the same answer. Use whichever is easier — typically $y$-integration when the curve is most naturally written as $x$ in terms of $y$.
You try
Find the area enclosed by $y = x^{2}$, the $y$-axis, and the line $y = 4$ (in the first quadrant), by integrating with respect to $y$.
Rearrange: $y = x^{2} \Rightarrow x = \sqrt{y}$ (first quadrant). Integrate from $y = 0$ to $y = 4$.
$x = \sqrt{y} = y^{1/2}$
$A = \displaystyle\int_{0}^{4} y^{1/2}\,dy = \Big[\,\dfrac{2}{3}\,y^{3/2}\,\Big]_{0}^{4}$
$= \dfrac{2}{3}(8) - 0 = \dfrac{16}{3}$
$A = \dfrac{16}{3}$ sq units
$\dfrac{16}{3}$ sq units
Section 11 of 14
The $y = x^{n}$ ratio — $B : A = n : 1$
A lovely result. Take the curve $y = x^{n}$ for $n > 0$, between $x = 0$ and $x = 1$ (so it goes from $(0,0)$ to $(1,1)$). It divides the unit square into two regions:
Region $A$: between the curve and the $x$-axis.
Region $B$: between the curve and the $y$-axis (above the curve, inside the unit square).
Find $A$. Integrate $y = x^{n}$ with respect to $x$ from $0$ to $1$:
$A = \displaystyle\int_{0}^{1} x^{n}\,dx = \Big[\,\dfrac{x^{n+1}}{n+1}\,\Big]_{0}^{1} = \dfrac{1}{n+1} - 0$
$A = \dfrac{1}{n+1}$
Find $B$. The unit square has area $1$, so:
$B = 1 - A = 1 - \dfrac{1}{n+1} = \dfrac{n+1 - 1}{n+1} = \dfrac{n}{n+1}$
The ratio:
$\dfrac{B}{A} = \dfrac{\;\dfrac{n}{n+1}\;}{\;\dfrac{1}{n+1}\;} = \dfrac{n}{n+1} \times \dfrac{n+1}{1} = n$
The result
$B : A = n : 1$
For $y = x^{n}$ between $0$ and $1$, the region above the curve (inside the unit square) is $n$ times the region below it.
Check. For $y = x$ ($n = 1$): $B : A = 1 : 1$. Diagonal of a square — each triangle is half. ✓
For $y = x^{2}$ ($n = 2$): $B : A = 2 : 1$. Below the parabola is $\tfrac{1}{3}$, above is $\tfrac{2}{3}$. ✓
For $y = x^{2}$ ($n = 2$): $B : A = 2 : 1$. Below the parabola is $\tfrac{1}{3}$, above is $\tfrac{2}{3}$. ✓
Equivalently, you can derive $B$ directly by integrating with respect to $y$. Since $y = x^{n}$ gives $x = y^{1/n}$:
$B = \displaystyle\int_{0}^{1} y^{1/n}\,dy = \Big[\,\dfrac{n\, y^{(n+1)/n}}{n+1}\,\Big]_{0}^{1} = \dfrac{n}{n+1}$ ✓
You try
Find the area between $y = x^{5}$ and the $y$-axis, from $y = 0$ to $y = 1$.
Apply $B : A = n : 1$ with $n = 5$. Since $A + B = 1$, $B = \dfrac{n}{n+1} = \dfrac{5}{6}$.
Direct method: $x = y^{1/5}$, so $B = \displaystyle\int_{0}^{1} y^{1/5}\,dy$
$= \Big[\,\dfrac{5\,y^{6/5}}{6}\,\Big]_{0}^{1} = \dfrac{5}{6}$
By the formula: $B = \dfrac{n}{n+1} = \dfrac{5}{6}$ ✓
$B = \dfrac{5}{6}$ sq unit
$\dfrac{5}{6}$ sq unit
Section 12 of 14
Application — motion: distance, velocity, acceleration
Velocity is the rate of change of distance. Acceleration is the rate of change of velocity. So integration takes you the other way — from acceleration back to velocity, from velocity back to distance.
Must learn — the chain
Differentiate down the chain: distance $\xrightarrow{\;\dfrac{d}{dt}\;}$ velocity $\xrightarrow{\;\dfrac{d}{dt}\;}$ acceleration
Integrate back up: acceleration $\xrightarrow{\;\int dt\;}$ velocity $\xrightarrow{\;\int dt\;}$ distance
Integrate back up: acceleration $\xrightarrow{\;\int dt\;}$ velocity $\xrightarrow{\;\int dt\;}$ distance
$s$ = distance (sometimes called displacement, $s(t)$). $v = \dfrac{ds}{dt}$. $a = \dfrac{dv}{dt} = \dfrac{d^{2}s}{dt^{2}}$.
Don't forget $+c$ — use the initial condition to find it.
Don't forget $+c$ — use the initial condition to find it.
Worked — a particle moves with velocity $v(t) = t^{2} - 6t + 1$ (in m/s). Initial displacement is $s = 5$ m. Find:
(a) the acceleration at $t = 3$ s
Acceleration is the derivative of velocity — we go down the chain. (Just for revision; the integration is in the next parts.)
$a(t) = \dfrac{dv}{dt} = 2t - 6$
$a(3) = 2(3) - 6 = 0$ m/s²
(b) the displacement function $s(t)$
Integrate the velocity. Use the initial condition $s(0) = 5$ to pin down $c$.
$s(t) = \displaystyle\int (t^{2} - 6t + 1)\,dt$
$= \dfrac{t^{3}}{3} - 3t^{2} + t + c$
At $t = 0$: $s(0) = 0 - 0 + 0 + c = 5$, so $c = 5$.
$s(t) = \dfrac{t^{3}}{3} - 3t^{2} + t + 5$
(c) the displacement at $t = 3$ s
$s(3) = \dfrac{27}{3} - 27 + 3 + 5 = 9 - 27 + 3 + 5 = -10$ m
Negative: the particle is $10$ m to the left of its starting point.
You try
A particle moves with acceleration $a(t) = t + 5$ m/s². It starts from rest ($v(0) = 0$) at the origin ($s(0) = 0$). Find the velocity at $t = 4$ and the distance travelled in the first $4$ seconds.
$v = \int a\,dt$, use $v(0) = 0$ for $c_{1}$. Then $s = \int v\,dt$, use $s(0) = 0$ for $c_{2}$.
$v(t) = \displaystyle\int (t + 5)\,dt = \dfrac{t^{2}}{2} + 5t + c_{1}$
$v(0) = 0 \Rightarrow c_{1} = 0$, so $v(t) = \dfrac{t^{2}}{2} + 5t$
$v(4) = 8 + 20 = 28$ m/s
$s(t) = \displaystyle\int \left(\dfrac{t^{2}}{2} + 5t\right) dt = \dfrac{t^{3}}{6} + \dfrac{5t^{2}}{2} + c_{2}$
$s(0) = 0 \Rightarrow c_{2} = 0$
$s(4) = \dfrac{64}{6} + 40 = \dfrac{32}{3} + 40 = \dfrac{32 + 120}{3} = \dfrac{152}{3}$ m
$v(4) = 28$ m/s · $s(4) = \dfrac{152}{3}$ m
$v(4) = 28$ m/s · $s(4) = \dfrac{152}{3}$ m
Section 13 of 14
Average value of a function
A function changes as you move through its domain. What's its average value over an interval $[a, b]$?
Idea: total accumulated value divided by length of interval. The total is the integral.
Must learn — average value
$\bar{f} = \dfrac{1}{b - a}\displaystyle\int_{a}^{b} f(x)\,dx$
Find the area, then divide by the width. The constant value that would give the same total.
Worked — ball trajectory: height (in metres) is $h(x) = 5x - x^{2}$ where $x$ is horizontal distance. Find the average height over the flight.
Find the flight range. Ball is on the ground when $h = 0$:
$5x - x^{2} = 0 \;\Rightarrow\; x(5 - x) = 0 \;\Rightarrow\; x = 0$ or $x = 5$
So the flight is from $x = 0$ to $x = 5$.
Apply the formula with $a = 0$, $b = 5$:
$\bar{h} = \dfrac{1}{5 - 0}\displaystyle\int_{0}^{5} (5x - x^{2})\,dx$
$= \dfrac{1}{5}\Big[\,\dfrac{5x^{2}}{2} - \dfrac{x^{3}}{3}\,\Big]_{0}^{5}$
$= \dfrac{1}{5}\left[\left(\dfrac{125}{2} - \dfrac{125}{3}\right) - 0\right]$
$= \dfrac{1}{5} \cdot \dfrac{375 - 250}{6} = \dfrac{1}{5} \cdot \dfrac{125}{6}$
$\bar{h} = \dfrac{25}{6}$ m ($\approx 4.17$ m)
Think of it as: a rectangle of width $5$ and height $\tfrac{25}{6}$ has the same area as the region under the parabola.
You try
A particle has velocity $v(t) = 3t^{2} - 18t + 15$ m/s, starting at rest at $t = 0$. Find the times when it is momentarily at rest, and the average velocity between those two times.
At rest when $v = 0$. Factor $3(t^{2} - 6t + 5) = 3(t - 1)(t - 5) = 0$, so $t = 1, 5$. Average over $[1, 5]$.
$3t^{2} - 18t + 15 = 0 \Rightarrow 3(t-1)(t-5) = 0$, so $t = 1$ and $t = 5$ s.
$\bar{v} = \dfrac{1}{5 - 1}\displaystyle\int_{1}^{5} (3t^{2} - 18t + 15)\,dt$
$= \dfrac{1}{4}\Big[\,t^{3} - 9t^{2} + 15t\,\Big]_{1}^{5}$
$= \dfrac{1}{4}\big[(125 - 225 + 75) - (1 - 9 + 15)\big]$
$= \dfrac{1}{4}\big[-25 - 7\big] = \dfrac{-32}{4} = -8$
At rest at $t = 1, 5$. · $\bar{v} = -8$ m/s
$t = 1, 5$ · $\bar{v} = -8$ m/s
Section 14 of 14
That's all three parts
You've now covered the complete Higher Level Integration syllabus across three lessons:
• Part 1 — Foundations & Trapezoidal Rule · anti-derivative, $+c$, area as a sum of slices
• Part 2 — The 8 Methods · everything that goes under the integral sign
• Part 3 — Definite Integrals & Areas · the long-question topic
• Part 2 — The 8 Methods · everything that goes under the integral sign
• Part 3 — Definite Integrals & Areas · the long-question topic
Exam tactics for the area long-question:
• Sketch first. Even a rough sketch. Where do the curves meet? What sits above what? Where does it cross the $x$-axis?
• Identify the bounded region clearly before writing any integral. Mark the limits and the orientation.
• If the region straddles the $x$-axis, split at the crossing point and take the modulus of each piece.
• For two-curve regions, find the intersection points (they're your limits) and integrate the difference, top minus bottom.
• If the question gives you a "show that" or "verify", work both sides carefully — the proof is the computation.
• Check your answer. If you can decompose it into rectangles and triangles, do so as a cross-check.
• Identify the bounded region clearly before writing any integral. Mark the limits and the orientation.
• If the region straddles the $x$-axis, split at the crossing point and take the modulus of each piece.
• For two-curve regions, find the intersection points (they're your limits) and integrate the difference, top minus bottom.
• If the question gives you a "show that" or "verify", work both sides carefully — the proof is the computation.
• Check your answer. If you can decompose it into rectangles and triangles, do so as a cross-check.
Integration is one of the biggest scoring opportunities on Paper 1. Get fluent here and the long question — typically worth around $25$ marks — is yours.
That's Integration.
From the anti-derivative, through the eight methods, to definite integrals and area-finding. The toolkit is complete — what's left is practice.