The area of one arc of y=sinx+cosx is
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Answered by
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Hello friend :-)
according to your question we need to find area of one arc of y => sinx+cosx
Now follow my steps what i am doing :=)
-------------------------------------------------------------------------------------------------------
see the figure it is the one arch of y => sinx +cosx
now we will do integration to calculate area
in the graph red line is for cos x and blue line is for sinx
To find the shaded area we could calculate the area under the graph of y =
sin x for x between 0 and 1 4π, and subtract this from the area under the
graph of y = cos x between the same limits. Alternatively the two processes
can be combined into one and we can write
π/4
shaded area => = ∫ (cos x − sin x)dx
0
π/4
=> {cosx + sinx }
0
=> cosπ/4 +sinπ/4 - cos0 - sin 0 = > 1/√2 + 1/√2 - 1
(2- √2 ) /√2=> => 0.414.
HOPE THIS HELPS YOU
@ engineer gopal IIT roorkey btech :=)
===========================================================
according to your question we need to find area of one arc of y => sinx+cosx
Now follow my steps what i am doing :=)
-------------------------------------------------------------------------------------------------------
see the figure it is the one arch of y => sinx +cosx
now we will do integration to calculate area
in the graph red line is for cos x and blue line is for sinx
To find the shaded area we could calculate the area under the graph of y =
sin x for x between 0 and 1 4π, and subtract this from the area under the
graph of y = cos x between the same limits. Alternatively the two processes
can be combined into one and we can write
π/4
shaded area => = ∫ (cos x − sin x)dx
0
π/4
=> {cosx + sinx }
0
=> cosπ/4 +sinπ/4 - cos0 - sin 0 = > 1/√2 + 1/√2 - 1
(2- √2 ) /√2=> => 0.414.
HOPE THIS HELPS YOU
@ engineer gopal IIT roorkey btech :=)
===========================================================
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Answered by
5
Answer. (2- √2 ) /√2=> => 0.414.
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