Question

Find the total area of the region bounded by the parabola \( y^2 = 4x \) and the straight line \( y = x \).

• A. \( \frac{4}{3} \)
• B. \( \frac{8}{3} \)
• C. \( \frac{16}{3} \)
• D. \( 2 \)

Hint:

Save time on standard calculus formats by memorizing this shortcut: the area enclosed between a standard parabola \( y^2 = 4ax \) and a passing line \( y = mx \) always evaluates precisely to \( \frac{8}{3} \frac{a^2}{m^3} \).

Answer 1

The Correct Option is B

Concept: The area bounded between two intersecting curves \( y_1 = f(x) \) and \( y_2 = g(x) \) from \( x = a \) to \( x = b \) is calculated by integrating the difference between the upper curve and the lower curve: \[ \text{Area} = \int_{a}^{b} \left( y_{\text{upper}} - y_{\text{lower}} \right) \, dx \] Finding the accurate boundary limits requires solving the equations simultaneously to locate their physical intersection coordinates.

Step 1: Find the intersection points of both curves.
Substitute the line equation \( y = x \) directly into the parabola equation \( y^2 = 4x \): \[ x^2 = 4x \quad \Rightarrow \quad x^2 - 4x = 0 \quad \Rightarrow \quad x(x - 4) = 0 \] This yields the intersection x-coordinates at \( x = 0 \) and \( x = 4 \). The corresponding points are \( (0,0) \) and \( (4,4) \).

Step 2: Set up and evaluate the definite integral framework.
In the interval \( [0, 4] \), the parabola curve \( y = 2\sqrt{x} \) forms the upper boundary, while the line \( y = x \) forms the lower boundary. Set up the difference integration: \[ \text{Area} = \int_{0}^{4} \left( 2\sqrt{x} - x \right) \, dx = \int_{0}^{4} \left( 2x^{\frac{1}{2}} - x \right) \, dx \] Apply the standard integration power rules: \[ = \left[ 2 \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{x^2}{2} \right]_{0}^{4} = \left[ \frac{4}{3}x^{\frac{3}{2}} - \frac{x^2}{2} \right]_{0}^{4} \]

Step 3: Evaluate the definite integral limits.
Substitute the upper limit \( x = 4 \) into the integrated expression (the lower limit at 0 vanishes): \[ \text{Area} = \left( \frac{4}{3}(4)^{\frac{3}{2}} - \frac{4^2}{2} \right) = \left( \frac{4}{3}(8) - \frac{16}{2} \right) = \frac{32}{3} - 8 \] Take the common denominator to isolate the final fractional area: \[ \text{Area} = \frac{32 - 24}{3} = \frac{8}{3} \text{ square units} \]