Write the curves in a convenient form

The given parabolas are: $y^2=2x-1 \,\Rightarrow\, x=\frac{y^2+1}{2}$ $y^2=4x-3 \,\Rightarrow\, x=\frac{y^2+3}{4}$

Since both are expressed naturally as $x$ in terms of $y$, it is easiest to integrate with respect to $y$.

Find the points of intersection

At intersection points, both expressions for $x$ are equal: $\frac{y^2+1}{2}=\frac{y^2+3}{4}$

Multiply by $4$: $2(y^2+1)=y^2+3$ $2y^2+2=y^2+3$ $y^2=1$ $y=\pm 1$

Now find corresponding $x$: $x=\frac{1+1}{2}=1$

So the curves intersect at $(1,1)$ and $(1,-1)$.

Determine which curve is on the right

For a fixed $y$ between $-1$ and $1$: $x_1=\frac{y^2+1}{2}, \qquad x_2=\frac{y^2+3}{4}$

Compare: $x_1-x_2=\frac{y^2+1}{2}-\frac{y^2+3}{4} = \frac{2y^2+2-y^2-3}{4}=\frac{y^2-1}{4}$

For $-1<y<1$, we have $y^2<1$, so $x_1-x_2<0$ Hence, $x_2>x_1$

Therefore,

• right curve: $x=\dfrac{y^2+3}{4}$
• left curve: $x=\dfrac{y^2+1}{2}$

Set up the area integral

Area enclosed is: $A=\int_{-1}^{1} \left(\frac{y^2+3}{4}-\frac{y^2+1}{2}\right)dy$

Simplify the integrand:

Evaluate the integral

Match with the options

$\boxed{\frac{1}{3}}$

This corresponds to Option A.