The area bounded by the parabola $y = x^2$ and the line $y = x$ is
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There is a standard shortcut formula for the area bounded between a parabola $x^2 = 4ay$ and a line $y = mx$: $\text{Area} = \frac{8a^2}{3m^3}$. Here, $4a = 1 \implies a = \frac{1}{4}$ and $m = 1$. Area = $\frac{8(1/16)}{3(1)} = \frac{1/2}{3} = \frac{1}{6}$.
Step 1: Know the two shapes.
We have a parabola $y=x^2$ (a U shape) and a straight line $y=x$. We want the area trapped between them.
Step 2: Find where they meet.
Set the two $y$ values equal: \[ x^2=x\Rightarrow x^2-x=0\Rightarrow x(x-1)=0. \] So they cross at $x=0$ and $x=1$. These are our limits.
Step 3: Decide which curve is on top.
Pick a value between $0$ and $1$, say $x=0.5$. The line gives $0.5$ and the parabola gives $0.25$. So the line $y=x$ is above the parabola in this region.
Step 4: Write the area as an integral.
Area between curves is the integral of (top minus bottom): \[ \text{Area}=\int_0^1\left(x-x^2\right)dx. \]
Step 5: Do the integration.
\[ =\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1. \]
Step 6: Put in the limits.
At $x=1$: $\frac{1}{2}-\frac{1}{3}$. At $x=0$: $0$. So \[ \text{Area}=\frac{1}{2}-\frac{1}{3}. \]
Step 7: Subtract the fractions.
Use a common bottom of $6$: \[ \frac{3}{6}-\frac{2}{6}=\frac{1}{6}. \] So the area is $\frac{1}{6}$ sq. units, which is option (4).
\[ \boxed{\dfrac{1}{6}\ \text{sq. units}} \]