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Question

Question: Area bounded by the curves y = sinx, tangent drawn to it at x = 0 and the line x = \(\frac { \pi } {...

Area bounded by the curves y = sinx, tangent drawn to it at x = 0 and the line x = π2\frac { \pi } { 2 }, is equal to

A

π242\frac { \pi ^ { 2 } - 4 } { 2 } sq. units

B

π244\frac { \pi ^ { 2 } - 4 } { 4 }sq. units

C

π224\frac { \pi ^ { 2 } - 2 } { 4 }sq. units

D

π222\frac { \pi ^ { 2 } - 2 } { 2 }sq. units

Answer

π244\frac { \pi ^ { 2 } - 4 } { 4 }sq. units

Explanation

Solution

The tangent drawn to y = sin x at x = 0 is the line y = x. Clearly the line y = x lies above the graph of y = sin x x(0,π2)\forall x \in \left( 0 , \frac { \pi } { 2 } \right).

Thus required area Δ=0π/2(xsinx)dx\Delta = \int _ { 0 } ^ { \pi / 2 } ( x - \sin x ) d x

=(x22+cosx)0π/2=π244= \left( \frac { x ^ { 2 } } { 2 } + \cos x \right) _ { 0 } ^ { \pi / 2 } = \frac { \pi ^ { 2 } - 4 } { 4 } sq. units.