`y = cos(x), y = sin(2x), x = 0, x = pi/2` Sketch the region enclosed by the given curves and find its area.
- print Print
- list Cite
Expert Answers
gsarora17
| Certified Educator
calendarEducator since 2015
write762 answers
starTop subjects are Math, Science, and Business
`y=cos(x) , y=sin(2x) , x=0 , x=pi/2`
Refer the attached image, y=cos(x) is plotted in red color and y=sin(2x) is plotted in blue color.
From graph,
cos(x) is above sin(2x) from 0 to pi/6
sin(2x) is above cos(x) fron pi6 to pi/2
Area of the region enclosed by the given curves A=`int_0^(pi/6)(cos(x)-sin(2x))dx+int_(pi/6)^(pi/2)((sin(2x)-cos(x))dx`
`A=[sin(x)-(-cos(2x)/2)]_0^(pi/6)+[-cos(2x)/2-sin(x)]_(pi/6)^(pi/2)`
`A=[sin(x)+cos(2x)/2]_0^(pi/6)+[-cos(2x)/2-sin(x)]_(pi/6)^(pi/2)`
`A=(sin(pi/6)+cos(pi/3)/2)-(sin(0)+cos(0)/2)+(-cos(pi)/2-sin(pi/2))-(-cos(pi/3)/2-sin(pi/6))`
`A=(1/2+1/4)-(0+1/2)+(1/2-1)-(-1/4-1/2)`
`A=1/4-1/2+1/4+1/2`
`A=1/2`
Images:
This image has been Flagged as inappropriate
Click to unflag
Related Questions
- Sketch the region enclosed by the given curves. y = 4 cos 2x, y = 4 − 4 cos 2x, 0 ≤ x ≤ π/2...
- 1 Educator Answer
- `y = cos(x), y = 1 - cos(x), 0<=x<=pi` Sketch the region enclosed by the given curves...
- 1 Educator Answer
- Find the area of the following region y=sin x and y=sin 2x from x=0 x=pi/2
- 1 Educator Answer
- `x = y^4, y = sqrt(2 - x), y = 0` Sketch the region enclosed by the given curves and find...
- 1 Educator Answer
- Suppose that 0<c<pi/2. For what value of c is the area of the region enclosed by the curves...
- 1 Educator Answer