The area S 3 is equal to | TestHub

Passage / Comprehension

The graphs of $y = f (x)$ and $y = sin x$ intersect at $P (a, f (a)), Q (π, 0)$ and $R (2 π, 0)$ , where $0 < a < π$ . The graph of $y = f (x)$ lies below that of $y = sin x$ for $0 < x < a$ , above it for $a < x < π$ , and below it for $π < x < 2 π$ . Let $S_{1}, S_{2}, S_{3}$ be the areas bounded by the curves $y = f (x)$ and $y = sin x$ between $x = 0$ and $x = a$ ; between $x = a$ and $x = π$ ; and between $x = π$ and $x = 2 π$ respectively. It is given that $S = 1 - sin t + (t - 1) cos t$ for all $t \leq a$ represents the area bounded between $y = f (x)$ and $y = sin x$ from $x = 0$ to $x = t$ .

The area $S_{3}$ is equal to

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Solution:

For $t \leq a, S (t) = \int_{0}^{t} (sin x - f (x)) d x$ . Differentiate: $sin t - f (t) = \frac{d}{d t} [1 - sin t + (t - 1) cos t] =$ $- cos t + cos t - (t - 1) sin t = (1 - t) sin t$ , so $f (t) = t sin t$ and $f (x) = x sin x . f (x) = sin x$ gives $sin x (x - 1) = 0$ , i.e. $x = 1$ or $x = nπ$ ; the first intersection is at $a = 1. S_{3} = \int_{π}^{2 π} (sin x - x sin x) d x =$ $[- cos x + x cos x - sin x]_{π}^{2 π} = (2 π - 1) - (1 - π) = 3 π - 2$ .

Mathematics - Area Under Curves Question with Solution | TestHub

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