Case Study Questions Application of Integration for Class 12

MCQ Questions:

1. 1. What are the points of intersection of \( y = \sqrt{x} \) and \( y = x^2 \)?

   • \( x = 0 \) and \( x = 2 \)
   • \( x = 0 \) and \( x = 1 \)
   • \( x = -1 \) and \( x = 1 \)
   • \( x = 1 \) and \( x = 4 \)

   Solution: Set \( \sqrt{x} = x^2 \Rightarrow x = 0, x = 1 \)

2. 2. Which function lies above the other in the interval \( [0, 1] \)?

   • \( y = x^2 \) lies above \( y = \sqrt{x} \)
   • Both are equal
   • \( y = \sqrt{x} \) lies above \( y = x^2 \)
   • Cannot determine

   Solution: For \( 0 < x < 1 \), \( \sqrt{x} > x^2 \)

3. 3. What is the correct expression for the area between the curves?

   • \( \int_0^1 (\sqrt{x} – x^2) \, dx \)
   • \( \int_0^1 (x^2 – \sqrt{x}) \, dx \)
   • \( \int_0^1 (x^2 + \sqrt{x}) \, dx \)
   • \( \int_0^1 (\sqrt{x} + x^2) \, dx \)

   Solution: The area is given by \( \int_0^1 [\sqrt{x} – x^2] \, dx \)

4. 4. Evaluate the integral \( \int_0^1 \sqrt{x} \, dx \)

   • \( \dfrac{2}{3} \)
   • \( \dfrac{1}{2} \)
   • \( \dfrac{3}{2} \)
   • \( 1 \)

   Solution: \( \int_0^1 x^{1/2} \, dx = \left[ \dfrac{2}{3} x^{3/2} \right]_0^1 = \dfrac{2}{3} \)

5. 5. Find the total area enclosed between the curves \( y = \sqrt{x} \) and \( y = x^2 \) from \( x = 0 \) to \( x = 1 \)

   • \( \dfrac{1}{6} \)
   • \( \dfrac{2}{3} \)
   • \( \dfrac{1}{3} \)
   • \( \dfrac{5}{6} \)

   Solution: \[ \int_0^1 (\sqrt{x} – x^2) \, dx = \int_0^1 x^{1/2} \, dx – \int_0^1 x^2 \, dx = \dfrac{2}{3} – \dfrac{1}{3} = \dfrac{5}{6} \]