Question 8 of 9advanced🔧 ApplyMCQ1 mark

Area bounded by the curve y=x3y = x^3, the xx-axis and the ordinates x=−2x = -2 and x=1x = 1 is

Exercise: Miscellaneous Exercise on Chapter 8 | Q: 4 | (Page Page 7)
For More Understanding

Explanation

The curve y = x³ crosses the x-axis at x = 0. For x ∈ (-2, 0), the curve lies below the x-axis (since x³ < 0 for x < 0), and for x ∈ (0, 1), it lies above the x-axis. Following the method shown in Example 3 of the context, we split the integral at x = 0 and take absolute value for the portion below the x-axis. The total area equals the sum of both portions.

Solution Steps

  1. Step 1: Identify that y = x³ crosses x-axis at x = 0. For x < 0, y < 0 (below x-axis); for x > 0, y > 0 (above x-axis).

  2. Step 2: Split the area into two parts: Area = |∫₋₂⁰ x³ dx| + ∫₀¹ x³ dx

  3. Step 3: Calculate ∫ x³ dx = x⁴/4

  4. Step 4: Area below x-axis: |[x⁴/4]₋₂⁰| = |0 - 16/4| = |−4| = 4

  5. Step 5: Area above x-axis: [x⁴/4]₀¹ = 1/4 - 0 = 1/4

  6. Step 6: Total Area = 4 + 1/4 = 16/4 + 1/4 = 17/4

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