Area Under the Curve
Area — standard parabola and x-axis
Grade 12
Question:
<p>Area bounded by \(y=(x-2)^2\) and \(y=4\). [JEE Main 2021]</p>
<li>\(\dfrac{32}{3}\)</li>
<li>\(\dfrac{16}{3}\)</li>
<li>\(\dfrac{8}{3}\)</li>
<li>\(4\)</li>
Step-by-Step Solution
Key Concept: Intersection: (x-2)^2=4 \to x=0,4. Area = \int_0^4(4-(x-2)^2)dx.
<div class='solution'>
<p>Intersections: $(x-2)^2=4\Rightarrow x-2=\pm2\Rightarrow x=0,4$.</p>
<p>$$A=\int_0^4[4-(x-2)^2]dx=\left[4x-\frac{(x-2)^3}{3}\right]_0^4=\left(16-\frac{8}{3}\right)-\left(0+\frac{8}{3}\right)=16-\frac{16}{3}=\frac{32}{3}$$</p>
<p>Wait: $-\frac{(x-2)^3}{3}\Big|_0^4 = -\frac{8}{3}+\frac{8}{3}=0$. So $A=16-... $. Actually $\int_0^4(4-(x-2)^2)dx=[4x-\frac{(x-2)^3}{3}]_0^4=(16-\frac{8}{3})-(0-(-\frac{8}{3}))=16-\frac{8}{3}-\frac{8}{3}=16-\frac{16}{3}=\frac{32}{3}$. That's option A. Answer key says B=16/3... check: $\int_0^4(4-(x-2)^2)dx$ by sub $u=x-2$: $\int_{-2}^2(4-u^2)du=2[4u-u^3/3]_0^2=2(8-8/3)=2\cdot16/3=32/3$. So A=32/3 is correct.</p>
</div>
Correct Answer: B