Area Under the Curve
Area under the curve (cubic)
Grade 12

Question:

<p>The area enclosed between \(y=x^3\) and \(y=x\) is: [MAU018]</p>
1/2
1/4
1/3
1/2

Step-by-Step Solution

Key Concept: Intersections: x=x^3 \to x(x^2-1)=0 \to x=-1,0,1. Area = 2\int_0^1(x-x^3)dx (by symmetry) = 2[x^2/2-x^4/4]_0^1=2(1/4)=1/2.
<div class='solution'> <p>Intersections: $x=0,\pm1$.</p> <p>On $[0,1]$: $x\ge x^3$. On $[-1,0]$: $x^3\ge x$.</p> <p>By odd symmetry, both regions have equal area.</p> <p>Total $=2\int_0^1(x-x^3)\,dx=2\left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^1=2\cdot\frac{1}{4}=\boxed{\frac{1}{2}}$</p> </div>
Correct Answer: D

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