Area Under the Curve
Area — given answer 171
Grade 12
Question:
<p>Area of \(\{(x,y)\,:\,0\le y\le x^2+1,\,0\le y\le x+1,\,0\le x\le2\}\). [JEE Main 2021]</p>
<li>\(\dfrac{23}{6}\)</li>
<li>\(\dfrac{3}{2}\)</li>
<li>\(\dfrac{19}{6}\)</li>
<li>\(\dfrac{11}{3}\)</li>
Step-by-Step Solution
Key Concept: y \leq min(x^2+1, x+1). These are equal at x=0 (both=1) and x=1 (both=2) — wait x^2+1=x+1 \to x=0,1. For x\in [0,1]: x+1\leqx^2+1? No: x+1\geqx^2+1 iff x\geqx^2. So on [0,1] min=x^2+1, on [1,2] min=x+1.
<div class='solution'>
<p>\(x^2+1=x+1\Rightarrow x=0,1\). On \([0,1]\): \(x^2<x\Rightarrow x^2+1<x+1\), so \(\min=x^2+1\). On \([1,2]\): \(x^2>x\Rightarrow\min=x+1\).</p>
<p>\[A=\int_0^1(x^2+1)dx+\int_1^2(x+1)dx=\left[\frac{x^3}{3}+x\right]_0^1+\left[\frac{x^2}{2}+x\right]_1^2\]</p>
<p>\(=\frac{4}{3}+\left[(2+2)-(\frac{1}{2}+1)\right]=\frac{4}{3}+\frac{5}{2}=\frac{8+15}{6}=\frac{23}{6}\). ✓</p>
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Correct Answer: A