Two more worked examples on the area between two curves — one with a line and a parabola, and one “show that” question where the curves cross inside the region so you must split the integral. Try each one in your book first, then click Show solution to check every step.
The routine (same every time): sketch → solve $f(x)=g(x)$ for the limits → decide which curve is on top (test a point) → integrate $\displaystyle\int (\text{top}-\text{bottom})\,dx$. If the curves swap over inside the region, split at the crossing and add the pieces.
Example 1 — a line and a parabola
EXAMPLE 1
Find the area of the region bounded between $y = 2x^2 - 1$ and $y = x$.
The line $y=x$ (coral) sits above the parabola $y=2x^2-1$ (navy) between the two intersection points; the shaded region is the area we want.
Step 1 — intersections. Set the curves equal: $2x^2 - 1 = x \;\Rightarrow\; 2x^2 - x - 1 = 0 \;\Rightarrow\; (2x+1)(x-1)=0$. By the Null Factor Law, $x = -\tfrac{1}{2}$ or $x = 1$. These are the limits of integration.
Step 2 — which curve is on top? Test $x=0$ (between the limits): the line gives $y=0$, the parabola gives $y=-1$. The line is above, so top $-$ bottom $= x - (2x^2-1)$.
Step 3 — set up the integral.
$$A = \int_{-1/2}^{\,1} \bigl[\,x - (2x^2-1)\,\bigr]\,dx = \int_{-1/2}^{\,1} \bigl(-2x^2 + x + 1\bigr)\,dx.$$
⚠️ Watch the signs. The lower limit is negative, so the substituted value $F\!\left(-\tfrac{1}{2}\right) = -\tfrac{7}{24}$ is itself negative — and you are subtracting it, so it adds $\tfrac{7}{24}$. Mishandling that one sign is the most common slip here (it leads to $\tfrac{31}{24}$ instead of the correct $\tfrac{9}{8}$).
Example 2 — a “show that” with a cross-over
EXAMPLE 2
Show that the area of the region bounded by the graphs of $f(x) = x^3$ and $g(x) = x(x+2)$ is $\dfrac{37}{12}$ square units.
The curves meet at $x=-1,\,0,\,2$. On $[-1,0]$ the cubic $f$ (navy) is on top; on $[0,2]$ the parabola $g$ (coral) is on top — so we split at $x=0$.
Step 1 — intersections. Set $f(x)=g(x)$: $x^3 = x(x+2) = x^2 + 2x$, so
$$x^3 - x^2 - 2x = 0 \;\Rightarrow\; x(x^2 - x - 2) = 0 \;\Rightarrow\; x(x+1)(x-2) = 0.$$
By the Null Factor Law, $x = -1,\; 0,\; 2$.
Step 2 — which curve is on top? The curves cross at $x=0$, which is inside the region, so we must split there.
On $[-1,0]$, test $x=-\tfrac{1}{2}$: $f=-\tfrac{1}{8}$, $g=-\tfrac{3}{4}$ → the cubic $f$ is on top.
On $[0,2]$, test $x=1$: $f=1$, $g=3$ → the parabola $g$ is on top.
Step 3 — set up (top $-$ bottom on each piece).
$$A = \int_{-1}^{0}\!\bigl(x^3 - x^2 - 2x\bigr)\,dx \;+\; \int_{0}^{2}\!\bigl(-x^3 + x^2 + 2x\bigr)\,dx.$$
$A = \dfrac{37}{12}$ square units, as required. $\blacksquare$
💡 Why split? If you integrated $\int_{-1}^{2}(f-g)\,dx$ in one go, the two pieces would partly cancel (one is positive, one negative) and you would not get the true enclosed area. Whenever the curves cross inside the interval, split at every crossing and add the pieces so each region counts positively.