Asked by sam
Find the area of the region enclosed between y=2sin(x and y=3cos(x) from x=0 to x=0.4pi
Hint: Notice that this region consists of two parts.
Notice: I'm getting 1.73762 but apparently that is wrong.
Hint: Notice that this region consists of two parts.
Notice: I'm getting 1.73762 but apparently that is wrong.
Answers
Answered by
Mgraph
Find x where 2sin(x)=3cos(x) (div by cos)
2tan(x)=3
tan(x)=1.5
x=56.31degr
Area(from x=0 to x=56.31)=
(3sin(56.31)+2cos(56.31)-(3sin(0)+2cos(0))
=3.60555-2=1.60555
Area(from x=56.31 to x=72)=
(-2cos(72)-3sin(72))-(-2cos(56.31)-3sin(56.31))=-3.47120-(-3.60555)=0.13435
The total area=1.60555+0.13435=1.7399
2tan(x)=3
tan(x)=1.5
x=56.31degr
Area(from x=0 to x=56.31)=
(3sin(56.31)+2cos(56.31)-(3sin(0)+2cos(0))
=3.60555-2=1.60555
Area(from x=56.31 to x=72)=
(-2cos(72)-3sin(72))-(-2cos(56.31)-3sin(56.31))=-3.47120-(-3.60555)=0.13435
The total area=1.60555+0.13435=1.7399
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.