<p>The ratio of area under \(y=x^n\) to area under \(y=x^{1/n}\) in \([0,1]\) (both above x-axis) is: [MAU043]</p>
Step-by-Step Solution
Key Concept: \int_0^1 xⁿdx = 1/(n+1). \int_0^1 x^(1/n)dx = n/(n+1). They are complementary (reflections about y=x) — their areas sum to 1.
<div class='solution'>
<p>$\int_0^1 x^n dx=\frac{1}{n+1}$.</p>
<p>$\int_0^1 x^{1/n}dx=\frac{n}{n+1}$.</p>
<p>Sum $=\frac{1}{n+1}+\frac{n}{n+1}=1$ ✓ (complementary areas about the diagonal $y=x$).</p>
<p>Ratio = $\frac{1/(n+1)}{n/(n+1)}=\frac{1}{n}$. Both areas together fill the unit square.</p>
</div>
Correct Answer: ['C']