JEE Mains · Maths · STD 12 - 8. Application and integration
The area of the region enclosed by the curves \(y=x^2-4 x+4\) and \(y^2=16-8 x\) is :
- A \(\frac{8}{3}\)
- B \(\frac{4}{3}\)
- C 8
- D 5
Answer & Solution
Correct Answer
(A) \(\frac{8}{3}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \text { Area } & =\int_0^2\left(\sqrt{16-8 x}-\left(x^2-4 x+4\right)\right) d x \\ & \left.=\frac{-(16-8 x)^{3 / 2}}{12}-\frac{x^3}{3}+2 x^2+4 x\right]_0^2 \\ & =\frac{8}{3}\end{aligned}\)
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