<p>Area bounded by \(y=x^3-3x^2+2x\) and \(y=0\). [JEE Main 2021]</p>
Step-by-Step Solution
Key Concept: y=x(x-1)(x-2). Roots at x=0,1,2. Positive on (0,1), negative on (1,2). Area = 2 \cdot \int_0^1 x(x-1)(x-2)dx (by |...|).
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<p>$y=x(x-1)(x-2)$. On $[0,1]$: positive. On $[1,2]$: negative.</p>
<p>$A=\int_0^1 x(x-1)(x-2)\,dx + \left|\int_1^2 x(x-1)(x-2)\,dx\right|$</p>
<p>$\int_0^1(x^3-3x^2+2x)dx=\frac{1}{4}-1+1=\frac{1}{4}$.</p>
<p>By symmetry (sub $x\to2-x$): $\int_1^2(x^3-3x^2+2x)dx=-\frac{1}{4}$.</p>
<p>Total area $=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$. ✓</p>
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Correct Answer: A