<p>The area bounded by the parabola \(y=x^2-4x+3\) and the \(x\)-axis is: [MAU012]</p>
Step-by-Step Solution
Key Concept: Roots: x=1 and x=3. Area = \int_1^3|x^2-4x+3|dx = -\int_1^3(x^2-4x+3)dx (since parabola is below axis between roots).
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<p>Roots: $x^2-4x+3=(x-1)(x-3)=0\Rightarrow x=1,3$.</p>
<p>Between roots, $y<0$ (downward on $[1,3]$).</p>
<p>$$A=-\int_1^3(x^2-4x+3)\,dx=-\left[\frac{x^3}{3}-2x^2+3x\right]_1^3$$</p>
<p>$=-\left[\left(9-18+9\right)-\left(\frac{1}{3}-2+3\right)\right]=-\left[0-\frac{4}{3}\right]=\frac{4}{3}$</p>
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Correct Answer: D