Area between two curves: top minus bottom, finding intersection points, splitting where the curves cross, integrating in y when the region is easier to slice sideways, and regions bounded by tangents and axes (Extension 1 only in the 2024 syllabus, as Advanced no longer lists it)

Find where they meet. $x^2 = 2x$ gives $x = 0$ or $x = 2$.

The line is above the parabola between these points (at $x = 1$, the line is at $2$ and the parabola at $1$), so

The area is $\frac{4}{3}$ square units.

Marker's note: one mark for the limits and the correct order of subtraction, one for the integration and value.

Intersections

$x = x^3$ gives $x(x - 1)(x + 1) = 0$, so $x = -1, 0, 1$. The curves cross at the origin, so the region has two parts.

Which is on top

On $0 \leq x \leq 1$ the line is above the cubic ($x \geq x^3$). On $-1 \leq x \leq 0$ the cubic is above the line (at $x = -\frac{1}{2}$, $x^3 = -\frac{1}{8}$ is greater than $x = -\frac{1}{2}$).

Use symmetry

Both curves are odd functions, so the two parts have equal area:

Marker's note: one mark for the three intersection points, one for recognising the change of top curve, one for the area. Integrating $\left( x - x^3 \right)$ over $[-1, 1]$ in one piece gives $0$, which is the classic error.

The curves meet where $e^x = e^{-x}$, that is $x = 0$. For $x > 0$, $e^x > e^{-x}$, so $e^x$ is the upper curve on $0 \leq x \leq 1$.

The exact area is $e + \frac{1}{e} - 2$, which is approximately $1.086$ square units.

Marker's note: one mark for the limits, one for the primitive (watch the sign from $\frac{d}{dx} e^{-x}$), one for the value.

They cross where $\sin x = \cos x$, at $x = \frac{\pi}{4}$ in this interval. Cosine is on top before this point and sine after it.

Split the integral.

The first integral is $\Big[ \sin x + \cos x \Big]_0^{\pi/4} = \sqrt{2} - 1$. The second is $\Big[ -\cos x - \sin x \Big]_{\pi/4}^{\pi/2} = -1 + \sqrt{2}$.

Marker's note: one mark for the crossing point, one for the two-part setup, one for the final value.

Slice across, not up

The line is $x = 6 - y$ and the parabola is $x = y^2$. Integrating in $y$ avoids the parabola's two branches and the two-piece integral in $x$ that they would need.

Intersections

$y^2 = 6 - y$ gives $y^2 + y - 6 = 0$, so $(y + 3)(y - 2) = 0$ and $y = -3$ or $y = 2$.

Right curve minus left curve

At $y = 0$ the line is at $x = 6$ and the parabola at $x = 0$, so the line is on the right:

At $y = 2$: $12 - 2 - \frac{8}{3} = \frac{22}{3}$. At $y = -3$: $-18 - \frac{9}{2} + 9 = -\frac{27}{2}$. So

Marker's note: one mark for the choice of $y$ as the variable and the two functions, one for the limits, one for the integration, one for the value.

Tangent

$\frac{dy}{dx} = 2x = 2$ at $x = 1$, so the tangent is $y - 1 = 2(x - 1)$, that is $y = 2x - 1$.

Which curve is on top

The parabola lies above all its tangents, so on $0 \leq x \leq 1$ the parabola is above the tangent. The region runs from the $y$-axis ($x = 0$) to the point of tangency ($x = 1$).

Integrate the difference

Marker's note: one mark for the tangent, one for identifying the limits, one for the integrand (the difference is a perfect square), one for the area.

Intersection. $x = \frac{1}{x}$ gives $x^2 = 1$, so $x = 1$ for $x > 0$. The region lies on $1 \leq x \leq 2$.

Top curve. At $x = 2$ the line gives $2$ and the hyperbola gives $\frac{1}{2}$, so $y = x$ is on top.

The area is $\frac{3}{2} - \ln 2$ square units, about $0.807$.

Marker's note: one mark for the intersection and limits, one for the top curve, one for the primitive of $\frac{1}{x}$, one for the exact value.

Intersections

$\dfrac{2}{1 + x^2} = 1$ gives $1 + x^2 = 2$, so $x = \pm 1$.

Top curve

At $x = 0$ the curve gives $2$ and the line gives $1$, so the curve is on top between $x = -1$ and $x = 1$.

Integrate

Use $\int \frac{2}{1 + x^2} dx = 2\tan^{-1} x$:

At $x = 1$: $2 \times \frac{\pi}{4} - 1 = \frac{\pi}{2} - 1$. At $x = -1$: $-\frac{\pi}{2} + 1$. Subtracting,

Marker's note: one mark for the intersection points, one for the top curve, one for the inverse tangent primitive, one for substituting both limits, one for the exact area.