<p>Area bounded by \(y=x(x-1)(x-2)\) and \(y=0\) over \([0,2]\). [JEE Main 2020]</p>
Step-by-Step Solution
Key Concept: Roots at 0,1,2. Positive on (0,1), negative on (1,2). Each half has area 1/4. Total = 1/2.
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<p>$\int_0^1 x(x-1)(x-2)dx=\int_0^1(x^3-3x^2+2x)dx=[\frac{x^4}{4}-x^3+x^2]_0^1=\frac{1}{4}-1+1=\frac{1}{4}$.</p>
<p>By symmetry (sub $x\to2-x$): $\int_1^2=-\frac{1}{4}$. Total area $=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$. ✓</p>
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Correct Answer: A