The parabola \( y^2 = 4x \) intersects the circle \( x^2 + y^2 = 5 \) at two points. To find these points of intersection, substitute \( x = \frac{y^2}{4} \) into \( x^2 + y^2 = 5 \):
\[ \left( \frac{y^2}{4} \right)^2 + y^2 = 5. \]
Simplify:
\[ \frac{y^4}{16} + y^2 = 5. \]
Multiply through by 16:
\[ y^4 + 16y^2 = 80. \]
Let \( z = y^2 \), so:
\[ z^2 + 16z - 80 = 0. \]
Solve this quadratic equation:
\[ z = \frac{-16 \pm \sqrt{16^2 - 4(1)(-80)}}{2(1)} = \frac{-16 \pm \sqrt{256 + 320}}{2} = \frac{-16 \pm \sqrt{576}}{2}. \] \[ z = \frac{-16 \pm 24}{2}. \]
This gives \( z = 4 \) or \( z = -20 \) (discarded as \( z = y^2 \geq 0 \)).
Thus, \( y^2 = 4 \), so \( y = \pm 2 \). The points of intersection are \( (1, 2) \) and \( (1, -2) \).
The region to the left of the parabola is bounded by the circle \( x^2 + y^2 = 5 \), and we calculate the area of this smaller region.
The area of the circle segment is:
\[ A_{\text{segment}} = R^2 \sin^{-1} \left( \frac{d}{R} \right) - \frac{d}{2} \sqrt{R^2 - d^2}, \]
where \( R = \sqrt{5} \) is the radius of the circle and \( d = 2 \) is the distance from the circle center to the chord.
Substitute the values:
\[ A_{\text{segment}} = 5 \sin^{-1} \left( \frac{2}{\sqrt{5}} \right) - \frac{2}{2} \sqrt{5 - 4}. \]
The area of the parabola sector is:
\[ A_{\text{parabola}} = \frac{2}{3}. \]
Thus, the smaller part’s area is:
\[ A_{\text{smaller}} = \frac{2}{3} + 5 \sin^{-1} \left( \frac{2}{\sqrt{5}} \right). \]
Thus the correct answer is Option 1.