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UPSC NDA I 2015 Mathematics Paper

Question Numbers: 78-79

For the next two (2) items that follow:

Consider the curves

y = sin x and y

= cos x

Question : 79 of 120

Marks: +1, -0

x=\frac{\pi}{4}

=\frac{\pi}{2}

Solution:

Concept:

Area between two curves is found by integrating the upper curve minus the lower curve over the given interval.

Explanation:

The two curves are

y = \sin x

and

y = \cos x

The region is bounded by

x = \frac{\pi}{4}

and

x = \frac{\pi}{2}

In this interval,

\sin x \ge \cos x

Thus, the area

A

is:

A = \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\sin x - \cos x) \, dx

Integrate term by term:

\int \sin x \, dx = -\cos x

\int \cos x \, dx = \sin x

A = \left[ -\cos x - \sin x \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}

Simplify:

A = - \left[ \cos x + \sin x \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}

Evaluate at

x = \frac{\pi}{2}

\cos\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{2}\right) = 0 + 1 = 1

x = \frac{\pi}{4}

\cos\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}

Therefore,

A = -[1 - \sqrt{2}] = \sqrt{2} - 1

Answer:

\sqrt{2} - 1

(Option A).

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