<p>The area bounded by \(y=|x-1|+|x-3|\) and the \(x\)-axis between \(x=0\) and \(x=4\) is: [MAU011]</p>
Step-by-Step Solution
Key Concept: Break into regions: [0,1], [1,3], [3,4]. |x-1|+|x-3| = (1-x)+(3-x)=4-2x on [0,1]; =2 on [1,3]; =(x-1)+(x-3)=2x-4 on [3,4].
<div class='solution'>
<p>On $[0,1]$: $y=(1-x)+(3-x)=4-2x$.</p>
<p>On $[1,3]$: $y=(x-1)+(3-x)=2$.</p>
<p>On $[3,4]$: $y=(x-1)+(x-3)=2x-4$.</p>
<p>$$A=\int_0^1(4-2x)dx+\int_1^3 2\,dx+\int_3^4(2x-4)dx$$</p>
<p>$=[4x-x^2]_0^1+[2x]_1^3+[x^2-4x]_3^4$</p>
<p>$=3+4+(0-(-3))=3+4+3=\boxed{10}$</p>
</div>
Correct Answer: A