Understanding the Area Bounded by Parabola $y^2=4x$ and Line $x=1$

This question asks us to find the area enclosed between the curve of the parabola $y^2 = 4x$ and the vertical line $x = 1$. This is a common problem in calculus involving finding the area between curves.

Finding Intersection Points

To set up the integral correctly, we first find the points where the parabola and the line intersect. We substitute $x = 1$ into the equation of the parabola:

• Equation of parabola: $y^2 = 4x$
• Substitute $x=1$: $y^2 = 4(1)$
• $y^2 = 4$
• Solving for $y$: $y = \pm \sqrt{4}$ which gives $y = 2$ and $y = -2$.

The intersection points are $(1, 2)$ and $(1, -2)$. The region is bounded on the right by $x=1$ and extends to the left, touching the y-axis at $x=0$ where the parabola's vertex is located.

Setting up the Area Integral

We can find the area using a definite integral. The parabola $y^2 = 4x$ can be expressed as $y = \pm 2\sqrt{x}$. This means the upper boundary of the region is $y = 2\sqrt{x}$ and the lower boundary is $y = -2\sqrt{x}$.

The area ($A$) is calculated by integrating the difference between the upper and lower functions with respect to $x$, over the interval from $x=0$ to $x=1$.

The area integral is set up as:

$A = \int_{0}^{1} (\text{Upper curve} - \text{Lower curve}) \, dx$

$A = \int_{0}^{1} (2\sqrt{x} - (-2\sqrt{x})) \, dx$

Simplifying the integrand:

$A = \int_{0}^{1} (2\sqrt{x} + 2\sqrt{x}) \, dx$

$A = \int_{0}^{1} 4\sqrt{x} \, dx$

Evaluating the Integral for Area

To find the area, we evaluate the definite integral found in the previous step. We rewrite $\sqrt{x}$ as $x^{1/2}$ to use the power rule for integration ($\int x^n dx = \frac{x^{n+1}}{n+1}$).

$A = 4 \int_{0}^{1} x^{1/2} \, dx$

Applying the power rule:

$A = 4 \left[ \frac{x^{1/2 + 1}}{1/2 + 1} \right]_{0}^{1}$

$A = 4 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1}$

$A = 4 \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1}$

Now, we substitute the limits of integration (upper limit $x=1$ and lower limit $x=0$):

$A = 4 \left( \frac{2}{3} (1)^{3/2} - \frac{2}{3} (0)^{3/2} \right)$

$A = 4 \left( \frac{2}{3} (1) - \frac{2}{3} (0) \right)$

$A = 4 \left( \frac{2}{3} - 0 \right)$

Final Area Calculation

Multiplying the constant by the result:

$A = 4 \times \frac{2}{3}$

$A = \frac{8}{3}$

Therefore, the area of the region bounded by the parabola $y^2 = 4x$ and the line $x = 1$ is $\frac{8}{3}$ square units.