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JEE Main 202522 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
22 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2025 · 22 January · Evening Shift · Mathematics · Area Under Curves

The area of the region enclosed by the curves y=x^2-4 x+4 and...

The area of the region enclosed by the curves y=x2−4x+4y=x^2-4 x+4 and y2=16−8xy^2=16-8 x is :

Options

  1. A83\frac{8}{3}
  2. B43\frac{4}{3}
  3. C8
  4. D5

Correct answer

A. 83\frac{8}{3}

Step-by-step solution

 Area =∫02(16−8x−(x2−4x+4))dx=−(16−8x)3/212−x33+2x2+4x]02=83\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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