Question

Area of the region bounded between $y^2 = 4x$ and $4x-2y-3=0$ is _______.

• A. $\int_{-1}^{3} (\frac{y}{2} + \frac{3}{4} + \frac{y^2}{4}) dy$
• B. $\int_{-1}^{3} (\frac{y}{2} + \frac{3}{4} - \frac{y^2}{4}) dy$
• C. $\int_{-1}^{3} (\frac{y^2}{4} - \frac{y}{2} - \frac{3}{4}) dy$
• D. None of these

Answer 1

Answer: (B)

B

Answer 2

To find the area of the region bounded by the given curves, we first need to find their intersection points. The given equations are:

1. $y^2 = 4x$ (Parabola)
2. $4x - 2y - 3 = 0$ (Line)

From equation (1), we can express $x$ in terms of $y$: $x = \frac{y^2}{4}$

Substitute this expression for $x$ into equation (2): $4\left(\frac{y^2}{4}\right) - 2y - 3 = 0$ $y^2 - 2y - 3 = 0$

This is a quadratic equation in $y$. We can factor it: $(y - 3)(y + 1) = 0$

This gives us the y-coordinates of the intersection points: $y = 3$ or $y = -1$.

Now, we find the corresponding x-coordinates using $x = \frac{y^2}{4}$: For $y = 3$: $x = \frac{3^2}{4} = \frac{9}{4}$. So, one intersection point is $(\frac{9}{4}, 3)$. For $y = -1$: $x = \frac{(-1)^2}{4} = \frac{1}{4}$. So, the other intersection point is $(\frac{1}{4}, -1)$.

To find the area between the curves, it's convenient to integrate with respect to $y$ since $x$ is given as a function of $y$ for the parabola, and can easily be expressed as a function of $y$ for the line. From the line equation $4x - 2y - 3 = 0$, we express $x$ in terms of $y$: $4x = 2y + 3$ $x = \frac{2y + 3}{4} = \frac{y}{2} + \frac{3}{4}$

So we have: $x_{\text{parabola}} = \frac{y^2}{4}$ $x_{\text{line}} = \frac{y}{2} + \frac{3}{4}$

To determine which curve is to the right (has a larger x-value) in the interval $y \in [-1, 3]$, we can pick a test value, say $y=0$: For $y=0$: $x_{\text{parabola}} = \frac{0^2}{4} = 0$ $x_{\text{line}} = \frac{0}{2} + \frac{3}{4} = \frac{3}{4}$

Since $\frac{3}{4} > 0$, the line $x_{\text{line}}$ is to the right of the parabola $x_{\text{parabola}}$ in the interval $[-1, 3]$.

The area $A$ of the region bounded by the curves is given by the integral of the difference between the right curve and the left curve, with respect to $y$, from the lower y-limit to the upper y-limit: $A = \int_{y_1}^{y_2} (x_{\text{right}} - x_{\text{left}}) dy$ $A = \int_{-1}^{3} \left( \left(\frac{y}{2} + \frac{3}{4}\right) - \left(\frac{y^2}{4}\right) \right) dy$ $A = \int_{-1}^{3} \left( \frac{y}{2} + \frac{3}{4} - \frac{y^2}{4} \right) dy$

Comparing this expression with the given options, the correct integral setup is given by option B.