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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=\cos 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

A. kk

Step-by-step solution

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

Let the required area be A=∫π/4π/2cos⁡x dxA = \int_{\pi/4}^{\pi/2} \cos x \, dx

Substituting x=π2−tx = \dfrac{\pi}{2} - t, we get dx=−dtdx = -dt

When x=π4x = \dfrac{\pi}{4}, t=π4t = \dfrac{\pi}{4} and when x=π2x = \dfrac{\pi}{2}, t=0t = 0

A=∫π/40cos⁡(π2−t)(−dt)A = \int_{\pi/4}^{0} \cos\left(\dfrac{\pi}{2} - t\right) (-dt)

A=∫0π/4sin⁡t dtA = \int_{0}^{\pi/4} \sin t \, dt

A=kA = k

Answer: kk
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