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NDA 2026
Exam
NDA
Year
2026
Subject
Mathematics
Chapter
Area Under Curves

NDA 2026 · Mathematics · Area Under Curves

Passage: Let k be the area between the curve y= x and x-axis in the...

Passage: Let kk be the area between the curve y=sin⁡xy=\sin x and x-axis in the interval \left[0,\dfrac{\pi}{4}\right].<br><br><strong>Question:</strong>Whatistheareabetweenthecurve.<br><br><strong>Question:</strong> What is the area between the curvey=\sin xandthex−axisintheinterval[π4,π2]and the x-axis in the interval \left[\dfrac{\pi}{4},\dfrac{\pi}{2}\right] ?

Options

  1. Akk
  2. B1−k1-k
  3. C(π−k)2\dfrac{(\pi-k)}{2}
  4. D(π−2k)2\dfrac{(\pi-2k)}{2}

Correct answer

B. 1−k1-k

Step-by-step solution

Given k=∫0π/4sin⁡x dxk = \int_{0}^{\pi/4} \sin x \, dx

k=[−cos⁡x]0π/4=−cos⁡(π4)+cos⁡(0)=1−12k = [-\cos x]_{0}^{\pi/4} = -\cos\left(\dfrac{\pi}{4}\right) + \cos(0) = 1 - \dfrac{1}{\sqrt{2}}

The required area is A=∫π/4π/2sin⁡x dxA = \int_{\pi/4}^{\pi/2} \sin x \, dx

A=[−cos⁡x]π/4π/2=−cos⁡(π2)+cos⁡(π4)=12A = [-\cos x]_{\pi/4}^{\pi/2} = -\cos\left(\dfrac{\pi}{2}\right) + \cos\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}

From the expression for kk, we have 12=1−k\dfrac{1}{\sqrt{2}} = 1 - k

Therefore, A=1−kA = 1 - k

Answer: 1−k1-k
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