Question

The area bounded by the curves \( y = x^2 \) and \( y = 2x \) in the first quadrant, is equal to:

• A. \( \frac{2}{3} \)
• B. \( \frac{4}{3} \)
• C. \( \frac{1}{3} \)
• D. \( \frac{8}{3} \)
• E. \( \frac{7}{3} \)

Hint:

When finding the area between two curves, subtract the lower curve from the upper curve and integrate over the region where the curves intersect. The limits of integration are determined by the points of intersection.

Answer 1

The Correct Option is B

We are asked to find the area bounded by the curves \( y = x^2 \) and \( y = 2x \) in the first quadrant.
Step 1: First, find the points of intersection of the two curves. Set the equations equal to each other:
\[ x^2 = 2x \] This simplifies to: \[ x^2 - 2x = 0 \] \[ x(x - 2) = 0 \]
Thus, the solutions are \( x = 0 \) and \( x = 2 \). These are the points of intersection of the curves in the first quadrant.
Step 2: The area between the curves from \( x = 0 \) to \( x = 2 \) is given by the integral:
\[ A = \int_0^2 \left( 2x - x^2 \right) \, dx \] Step 3: Now, evaluate the integral: \[ A = \int_0^2 2x \, dx - \int_0^2 x^2 \, dx \] \[ A = \left[ x^2 \right]_0^2 - \left[ \frac{x^3}{3} \right]_0^2 \] \[ A = (2^2 - 0^2) - \left( \frac{2^3}{3} - \frac{0^3}{3} \right) \] \[ A = 4 - \frac{8}{3} \] \[ A = \frac{12}{3} - \frac{8}{3} = \frac{4}{3} \]
Thus, the area bounded by the curves is \( \frac{4}{3} \).
Therefore, the correct answer is option (B).