If $f (x) = sin x, \forall x \in [0, \frac{π}{2}], f (x) + f (π - x) = 2, \forall x \in (\frac{π}{2}, π) a n d f (x) = f (2 π - x), \forall x \in (π, 2 π),$ then the area enclosed by $y = f (x)$ and the x-axis is $π s q \overset{u}{˙} n i t s$ (b) $2 π s q \overset{u}{˙} n i t s$ $2 s q \overset{u}{˙} n i t s$ (d) $4 s q \overset{u}{˙} n i t s$

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Benjamin Discussed

f (x) = sin x, \forall x \in [0, \frac{π}{2}], f (x) + f (π - x) = 2, \forall x \in (\frac{π}{2}, π) a n d f (x) = f (2 π - x), \forall x \in (π, 2 π),

then the area enclosed by

y = f (x)

and the x-axis is

π s q \overset{u}{˙} n i t s

(b)

2 π s q \overset{u}{˙} n i t s

2 s q \overset{u}{˙} n i t s

(d)

4 s q \overset{u}{˙} n i t s

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Question Text

f (x) = sin x, \forall x \in [0, \frac{π}{2}], f (x) + f (π - x) = 2, \forall x \in (\frac{π}{2}, π) a n d f (x) = f (2 π - x), \forall x \in (π, 2 π),

then the area enclosed by

y = f (x)

and the x-axis is

π s q \overset{u}{˙} n i t s

(b)

2 π s q \overset{u}{˙} n i t s

2 s q \overset{u}{˙} n i t s

(d)

4 s q \overset{u}{˙} n i t s

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