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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Area Under Curves

NDA 2017 · Mathematics · Area Under Curves

The area bounded by the curve |x|+|y|=1 is

The area bounded by the curve ∣x∣+∣y∣=1|x|+|y|=1 is

Options

  1. A11 square unit
  2. B222 \sqrt{2} square units
  3. C22 square unit
  4. D232 \sqrt{3} square units

Correct answer

C. 22 square unit

Step-by-step solution

∣x∣+∣y∣=1|x|+|y|=1 We know, |x|=\left\{x,x≥0−x,x<0\begin{array}{c}x, x \geq 0 \\ -x, x < 0\end{array}\right.\therefore|x|+|y|=1 is \left\{x+y=1 for  x>0,y>0−x+y=1 for  x<0,y>0x−y=1 for  x>0,y<0−x−y=1 for x<0,y<0\begin{array}{cc}x+y=1 & \text{\text{ for } } x > 0, y > 0 \\ -x+y=1 & \text{\text{ for } } x < 0, y > 0 \\ x-y=1 & \text{\text{ for } } x > 0, y < 0 \\ -x-y=1 & \text{ for } x < 0, y < 0\end{array}\right. If we plot graphs of these equations, we get The curve is symmetrical about x any y -axis. ∴ Area =4× Area of AOD \therefore \text{ Area }=4 \times \text{ Area of AOD }=4×∫01ydx=4×∫01(1−x)dx=4[x−x22]01=4[1−122−(0−022)]=4(1−12)=4×12=2\begin{aligned} & =4 \times \int_0^1 y d x=4 \times \int_0^1(1-x) d x=4\left[x-\frac{x^2}{2}\right]_0^1 \\ & =4\left[1-\frac{1^2}{2}-\left(0-\frac{0^2}{2}\right)\right] \\ & =4\left(1-\frac{1}{2}\right)=4 \times \frac{1}{2}=2\end{aligned}$$
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