Trig function help!! (1 Viewer)

[redpint]

Hi guys! How would you solve these questions?

1. find the exact area enclosed by the curve y = sinx and the line y=1/2 for 0≤ x ≤ 2∏

2. find the exact area bounded by the curves y =sinx and y=cosx in the domain 0≤ x ≤ 2∏.

Thank you in advance!

Answers Are
1. 3√3 - ∏ / 3 squared units
2. 2√2 squared units

 

[darlking]

For 1) SHouldn't the answer be r3 - pi/3

Area between the cuerve y=sinx and line y=1/2 under the curve =∫(sinx−1/2)dx
Y= sinx
When y= 0, sinx=0 and x=0
When y= 1/2, sinx=1/2 and x=π/6 and 5π/6 for 0 ≤ x ≤ 2pi
Hence Area between the cuerve y=sinx and line y=1/2 under the curve
=∫ from (π/6 to 5π/6) (sinx−1/2)dx
= (−cosx−1/2x) (π/6 to 5π/6)
= {−cos(5π/6)−1/2(5π/6) } −{−cos(π/6)−1/2(π/6)}
= {−(−√3/2)−(5π/12) } −{−(√3/2)−(π/12)}
= {√3/2−5π/12+√3/2+π/12}
= {√3−π/3} sq. units

 

[darlking]

2. find the exact area bounded by the curves y =sinx and y=cosx in the domain 0≤ x ≤ 2∏.

Find P.O.I when you let sinx=cosx becomes tanx=1. Solutions for x= 5pi/4 and pi/4.

A= integral between 5pi/4 and pi/4 (sinx-cosx)dx.
= -cosx-sinx between 5pi/4 and pi/4

=[ -cos5pi/4 - sin5pi/4 ] - [ -cospi/4 - sinpi/4 ]

simplify and rationalise the deno and u get 2 root 2