Question

The area of the region bounded by the curves \(y=x^3\), \(y=x^2\) and the lines \(x=0\) and \(x=2\) is

• A. \(\frac{4}{3}\)
• B. \(\frac{3}{2}\)
• C. \(\frac{2}{3}\)
• D. \(\frac{5}{3}\)

Hint:

Always check for intersection points within the integration limits. If curves cross, you must split the integral and take the absolute difference \(|f(x) - g(x)|\).

Answer 1

The Correct Option is B

Step 1: Identify Intersection Points:
Solve \(x^3 = x^2 \implies x^2(x-1) = 0\). Intersections at \(x=0\) and \(x=1\). The region is bounded by \(x=0\) and \(x=2\). So we split the integral at \(x=1\).
Step 2: Determine Upper Curve:
In interval \((0, 1)\): Test \(x=0.5\). \(y=x^2=0.25\), \(y=x^3=0.125\). So \(x^2>x^3\). In interval \((1, 2)\): Test \(x=1.5\). \(y=x^2=2.25\), \(y=x^3=3.375\). So \(x^3>x^2\).
Step 3: Calculate Area:
\[ A = \int_0^1 (x^2 - x^3) dx + \int_1^2 (x^3 - x^2) dx \] Area 1: \[ A_1 = \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_0^1 = \left( \frac{1}{3} - \frac{1}{4} \right) - 0 = \frac{1}{12} \] Area 2: \[ A_2 = \left[ \frac{x^4}{4} - \frac{x^3}{3} \right]_1^2 = \left( \frac{16}{4} - \frac{8}{3} \right) - \left( \frac{1}{4} - \frac{1}{3} \right) \] \[ A_2 = \left( 4 - \frac{8}{3} \right) - \left( \frac{3-4}{12} \right) = \frac{4}{3} - \left( -\frac{1}{12} \right) = \frac{16}{12} + \frac{1}{12} = \frac{17}{12} \] Total Area: \[ A = A_1 + A_2 = \frac{1}{12} + \frac{17}{12} = \frac{18}{12} = \frac{3}{2} \]