Question:

The area (in square units) of the region bounded by curves y=x y = x and y=x3 y = x^3 is:

Updated On: Mar 27, 2025
  • 0
  • 12 \frac{1}{2}
  • 14 \frac{1}{4}
  • 4
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The Correct Option is C

Solution and Explanation

The given curves are y=x y = x and y=x3 y = x^3 . To find the area of the region between these curves, first find the points of intersection:

x=x3    x(x21)=0    x=0 or x=±1. x = x^3 \implies x(x^2 - 1) = 0 \implies x = 0 \text{ or } x = \pm 1.

The region lies between x=0 x = 0 and x=1 x = 1 (since negative values will mirror the same area). The area between the curves is:

Area=01(xx3)dx. \text{Area} = \int_0^1 (x - x^3) \, dx.

Evaluate the integral:

01(xx3)dx=01xdx01x3dx. \int_0^1 (x - x^3) \, dx = \int_0^1 x \, dx - \int_0^1 x^3 \, dx.

Compute each term:

01xdx=[x22]01=12, \int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2}, 01x3dx=[x44]01=14. \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4}.

Subtract the results:

Area=1214=14. \text{Area} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}.

Thus, the area of the region is 14 \frac{1}{4} square units.

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