What is the area between the curve y= x and the x-axis in the interval ≤ft[π/4, π/2] ?

Correct Answer is : 1−k1-{k}1−k

Correct Answer is : 1−k1-{k}1−k

Test Index

UPSC NDA I 2026 Mathematics Paper

Question Numbers: 87-88

For the next two (02) items that follow:

Let k be the area between the curve

\mathrm{y}=\sin \mathrm{x}

and x-axis in the interval

\left[0, \frac{\pi}{4}\right]

Question : 87 of 120

Marks: +1, -0

y=\sin x

x

\left[\frac{\pi}{4}, \frac{\pi}{2}\right]

Solution:

Concept:

Area under

y = \sin x

is found by integrating

\sin x

between the given limits.

Explanation:

Area in

\left[0, \frac{\pi}{4}\right]

k = \int_{0}^{\pi/4} \sin x \, dx = 1 - \frac{\sqrt{2}}{2}

So,

\frac{\sqrt{2}}{2} = 1 - k

Required area in

\left[\frac{\pi}{4}, \frac{\pi}{2}\right]

\int_{\pi/4}^{\pi/2} \sin x \, dx = \frac{\sqrt{2}}{2}

\therefore \text{Required area} = 1 - k

Answer:

1 - k

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