Question

Find the area bounded between the parabola \( y^2 = 4x \) and the line \( y = 2x - 3 \).

• A. \( \displaystyle \int_{1 - \sqrt{7}}^{1 + \sqrt{7}} \left( \frac{y + 3}{2} - \frac{y^2}{4} \right) dy \)
• B. \( \displaystyle \int_{1 - \sqrt{7}}^{1 + \sqrt{7}} \left( \frac{y^2}{4} - \frac{y + 3}{2} \right) dy \)
• C. \( \displaystyle \int_{-2}^{2} \left( \frac{y^2}{4} - y \right) dy \)
• D. \( \displaystyle \int_{1 - \sqrt{7}}^{1 + \sqrt{7}} \left( \frac{y^2 + y + 3}{2} \right) dy \)

Hint:

When dealing with curves like \( y^2 = 4x \), consider expressing \( x \) in terms of \( y \) and integrating along the vertical strip when bounded by a function of \( y \).

Answer 1

The Correct Option is A

We are given: - Parabola: \( y^2 = 4x \Rightarrow x = \frac{y^2}{4} \) - Line: \( y = 2x - 3 \Rightarrow x = \frac{y + 3}{2} \) We will find the area between the curves by integrating the horizontal distance between the curves (i.e., \( x_{\text{line}} - x_{\text{parabola}} \)) with respect to \( y \). Step 1: Find points of intersection Equating the two expressions for \( x \): \[ \frac{y^2}{4} = \frac{y + 3}{2} \Rightarrow \frac{y^2}{4} - \frac{y + 3}{2} = 0 \Rightarrow \frac{y^2 - 2y - 6}{4} = 0 \Rightarrow y^2 - 2y - 6 = 0 \] Solve the quadratic: \[ y = \frac{2 \pm \sqrt{4 + 24}}{2} = \frac{2 \pm \sqrt{28}}{2} = \frac{2 \pm 2\sqrt{7}}{2} = 1 \pm \sqrt{7} \] So, limits of integration are from \( y = 1 - \sqrt{7} \) to \( y = 1 + \sqrt{7} \). Step 2: Setup the integral The area is: \[ A = \int_{1 - \sqrt{7}}^{1 + \sqrt{7}} \left( \frac{y + 3}{2} - \frac{y^2}{4} \right) dy \] Simplify the integrand: \[ \frac{y + 3}{2} - \frac{y^2}{4} = -\frac{y^2}{4} + \frac{y + 3}{2} \] Step 3: Integrate Break it into parts: \[ A = \int_{1 - \sqrt{7}}^{1 + \sqrt{7}} \left( -\frac{y^2}{4} + \frac{y}{2} + \frac{3}{2} \right) dy = -\frac{1}{4} \int y^2 dy + \frac{1}{2} \int y dy + \frac{3}{2} \int dy \] Evaluate each term: \[ \int y^2 dy = \frac{y^3}{3}, \quad \int y dy = \frac{y^2}{2}, \quad \int dy = y \] So: \[ A = \left[ -\frac{1}{4} \cdot \frac{y^3}{3} + \frac{1}{2} \cdot \frac{y^2}{2} + \frac{3}{2} y \right]_{1 - \sqrt{7}}^{1 + \sqrt{7}} = \left[ -\frac{y^3}{12} + \frac{y^2}{4} + \frac{3y}{2} \right]_{1 - \sqrt{7}}^{1 + \sqrt{7}} \] Now apply the identity for definite integrals over symmetric limits about a point \( a \), i.e., if \( f(y) \) is even around \( y = 1 \), define: Let \( y = 1 + t \Rightarrow \) symmetry around \( y = 1 \). It simplifies the calculation. Alternatively, compute numerically using symmetry: Since the integrand is even about \( y = 1 \), the area becomes: \[ A = 2 \int_0^{\sqrt{7}} \left( -\frac{(1 + t)^3}{12} + \frac{(1 + t)^2}{4} + \frac{3(1 + t)}{2} \right) dt \] But this becomes long — hence exact simplification is better done via substitution or numerical computation. Conclusion: The integral can be evaluated explicitly as shown, or you can evaluate the expression: \[ \boxed{A = \int_{1 - \sqrt{7}}^{1 + \sqrt{7}} \left( \frac{y + 3}{2} - \frac{y^2}{4} \right) dy} \] This gives the exact bounded area between the parabola and the line.