Area Under the Curve
Area using standard regions
Grade 12
Question:
<p>The area of the region \(R=\{(x,y)\,:\,x^2\le y\le |x|\}\) is: [MAU016]</p>
Step-by-Step Solution
Key Concept: By symmetry, consider x\geq0 where region is x^2\leqy\leqx. Area for x\in [0,1] = \int_0^1(x-x^2)dx = 1/6. Double by symmetry: 1/3.
<div class='solution'>
<p>For $x\ge0$: region is $x^2\le y\le x$, $x\in[0,1]$.</p>
<p>Area (right half) $=\int_0^1(x-x^2)dx=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$.</p>
<p>By symmetry ($y=(-x)^2=x^2$ and $y=|x|$ both symmetric about y-axis): total area $=2\cdot\frac{1}{6}=\boxed{\frac{1}{3}}$.</p>
</div>
Correct Answer: B