Area between the curves of $2cos(x)$ and $x/2$











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I'm trying to obtain the area between the curve of these two functions (for x>0), lets call them $f(x)=2cos(x)$ and $g(x)=x/2$ and my idea is to get the area under the curve of $f(x)$, then subtract the sum of these: the area under the curve of $g(x)$ [$0$, intersection point] and $f(x)$ [intersection point, $pi/2$]



Is this the right way or theres an easier way?



Thanks.






functions definite-integrals graphing-functions area curves







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  • 1




    Have you tried to plot the functions to get an understanding on how many times they intersect?
    – maxmilgram
    1 hour ago






  • 1




    I don't think you can find the intersections exactly. Even wolfram can't do it so I'm not even going to try :)
    – vrugtehagel
    1 hour ago












  • Yes i checked and only need x>0. And yes i checked the intersection is tricky but then how should i do it?
    – Cliff
    59 mins ago










  • Why do you only need $x>0$? Please update the problem description to give the full problem.
    – maxmilgram
    54 mins ago










  • get the area under the curve of f(x), then subtract the sum of these: the area under the curve of g(x) [0, intersection point] and ${color{red}{text{f(x) [intersection point, π/2]}}}$. Why are you doing the red part? Could you carefully retype your the question? Please specify if you're looking for an exact answer or if a numerical solution suffices.
    – zahbaz
    52 mins ago

















up vote
1
down vote

favorite












I'm trying to obtain the area between the curve of these two functions (for x>0), lets call them $f(x)=2cos(x)$ and $g(x)=x/2$ and my idea is to get the area under the curve of $f(x)$, then subtract the sum of these: the area under the curve of $g(x)$ [$0$, intersection point] and $f(x)$ [intersection point, $pi/2$]



Is this the right way or theres an easier way?



Thanks.






functions definite-integrals graphing-functions area curves







share|cite|improve this question




















  • 1




    Have you tried to plot the functions to get an understanding on how many times they intersect?
    – maxmilgram
    1 hour ago






  • 1




    I don't think you can find the intersections exactly. Even wolfram can't do it so I'm not even going to try :)
    – vrugtehagel
    1 hour ago












  • Yes i checked and only need x>0. And yes i checked the intersection is tricky but then how should i do it?
    – Cliff
    59 mins ago










  • Why do you only need $x>0$? Please update the problem description to give the full problem.
    – maxmilgram
    54 mins ago










  • get the area under the curve of f(x), then subtract the sum of these: the area under the curve of g(x) [0, intersection point] and ${color{red}{text{f(x) [intersection point, π/2]}}}$. Why are you doing the red part? Could you carefully retype your the question? Please specify if you're looking for an exact answer or if a numerical solution suffices.
    – zahbaz
    52 mins ago















up vote
1
down vote

favorite









up vote
1
down vote

favorite











I'm trying to obtain the area between the curve of these two functions (for x>0), lets call them $f(x)=2cos(x)$ and $g(x)=x/2$ and my idea is to get the area under the curve of $f(x)$, then subtract the sum of these: the area under the curve of $g(x)$ [$0$, intersection point] and $f(x)$ [intersection point, $pi/2$]



Is this the right way or theres an easier way?



Thanks.






functions definite-integrals graphing-functions area curves







share|cite|improve this question















I'm trying to obtain the area between the curve of these two functions (for x>0), lets call them $f(x)=2cos(x)$ and $g(x)=x/2$ and my idea is to get the area under the curve of $f(x)$, then subtract the sum of these: the area under the curve of $g(x)$ [$0$, intersection point] and $f(x)$ [intersection point, $pi/2$]



Is this the right way or theres an easier way?



Thanks.






functions definite-integrals graphing-functions area curves




functions definite-integrals graphing-functions area curves






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited 46 mins ago

























asked 1 hour ago









Cliff

84




84








  • 1




    Have you tried to plot the functions to get an understanding on how many times they intersect?
    – maxmilgram
    1 hour ago






  • 1




    I don't think you can find the intersections exactly. Even wolfram can't do it so I'm not even going to try :)
    – vrugtehagel
    1 hour ago












  • Yes i checked and only need x>0. And yes i checked the intersection is tricky but then how should i do it?
    – Cliff
    59 mins ago










  • Why do you only need $x>0$? Please update the problem description to give the full problem.
    – maxmilgram
    54 mins ago










  • get the area under the curve of f(x), then subtract the sum of these: the area under the curve of g(x) [0, intersection point] and ${color{red}{text{f(x) [intersection point, π/2]}}}$. Why are you doing the red part? Could you carefully retype your the question? Please specify if you're looking for an exact answer or if a numerical solution suffices.
    – zahbaz
    52 mins ago
















  • 1




    Have you tried to plot the functions to get an understanding on how many times they intersect?
    – maxmilgram
    1 hour ago






  • 1




    I don't think you can find the intersections exactly. Even wolfram can't do it so I'm not even going to try :)
    – vrugtehagel
    1 hour ago












  • Yes i checked and only need x>0. And yes i checked the intersection is tricky but then how should i do it?
    – Cliff
    59 mins ago










  • Why do you only need $x>0$? Please update the problem description to give the full problem.
    – maxmilgram
    54 mins ago










  • get the area under the curve of f(x), then subtract the sum of these: the area under the curve of g(x) [0, intersection point] and ${color{red}{text{f(x) [intersection point, π/2]}}}$. Why are you doing the red part? Could you carefully retype your the question? Please specify if you're looking for an exact answer or if a numerical solution suffices.
    – zahbaz
    52 mins ago










1




1




Have you tried to plot the functions to get an understanding on how many times they intersect?
– maxmilgram
1 hour ago




Have you tried to plot the functions to get an understanding on how many times they intersect?
– maxmilgram
1 hour ago




1




1




I don't think you can find the intersections exactly. Even wolfram can't do it so I'm not even going to try :)
– vrugtehagel
1 hour ago






I don't think you can find the intersections exactly. Even wolfram can't do it so I'm not even going to try :)
– vrugtehagel
1 hour ago














Yes i checked and only need x>0. And yes i checked the intersection is tricky but then how should i do it?
– Cliff
59 mins ago




Yes i checked and only need x>0. And yes i checked the intersection is tricky but then how should i do it?
– Cliff
59 mins ago












Why do you only need $x>0$? Please update the problem description to give the full problem.
– maxmilgram
54 mins ago




Why do you only need $x>0$? Please update the problem description to give the full problem.
– maxmilgram
54 mins ago












get the area under the curve of f(x), then subtract the sum of these: the area under the curve of g(x) [0, intersection point] and ${color{red}{text{f(x) [intersection point, π/2]}}}$. Why are you doing the red part? Could you carefully retype your the question? Please specify if you're looking for an exact answer or if a numerical solution suffices.
– zahbaz
52 mins ago






get the area under the curve of f(x), then subtract the sum of these: the area under the curve of g(x) [0, intersection point] and ${color{red}{text{f(x) [intersection point, π/2]}}}$. Why are you doing the red part? Could you carefully retype your the question? Please specify if you're looking for an exact answer or if a numerical solution suffices.
– zahbaz
52 mins ago












1 Answer
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I guess you could use Wolfram's approximation for the intersection point: $xapprox1.25235$.



Then you get $approxint_0^{1.25235} (2cos x-frac x2)operatorname dx=[2sin x-frac {x^2}4]_0^{1.25235}=2sin 1.25235-frac{(1.25235)^2}4approx1.50735$ for the area between the curves.






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  • Thanks my brain is messed up due to lack of sleep.
    – Cliff
    5 mins ago











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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes








up vote
0
down vote



accepted










I guess you could use Wolfram's approximation for the intersection point: $xapprox1.25235$.



Then you get $approxint_0^{1.25235} (2cos x-frac x2)operatorname dx=[2sin x-frac {x^2}4]_0^{1.25235}=2sin 1.25235-frac{(1.25235)^2}4approx1.50735$ for the area between the curves.






share|cite|improve this answer





















  • Thanks my brain is messed up due to lack of sleep.
    – Cliff
    5 mins ago















up vote
0
down vote



accepted










I guess you could use Wolfram's approximation for the intersection point: $xapprox1.25235$.



Then you get $approxint_0^{1.25235} (2cos x-frac x2)operatorname dx=[2sin x-frac {x^2}4]_0^{1.25235}=2sin 1.25235-frac{(1.25235)^2}4approx1.50735$ for the area between the curves.






share|cite|improve this answer





















  • Thanks my brain is messed up due to lack of sleep.
    – Cliff
    5 mins ago













up vote
0
down vote



accepted







up vote
0
down vote



accepted






I guess you could use Wolfram's approximation for the intersection point: $xapprox1.25235$.



Then you get $approxint_0^{1.25235} (2cos x-frac x2)operatorname dx=[2sin x-frac {x^2}4]_0^{1.25235}=2sin 1.25235-frac{(1.25235)^2}4approx1.50735$ for the area between the curves.






share|cite|improve this answer












I guess you could use Wolfram's approximation for the intersection point: $xapprox1.25235$.



Then you get $approxint_0^{1.25235} (2cos x-frac x2)operatorname dx=[2sin x-frac {x^2}4]_0^{1.25235}=2sin 1.25235-frac{(1.25235)^2}4approx1.50735$ for the area between the curves.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 19 mins ago









Chris Custer

8,5692623




8,5692623












  • Thanks my brain is messed up due to lack of sleep.
    – Cliff
    5 mins ago


















  • Thanks my brain is messed up due to lack of sleep.
    – Cliff
    5 mins ago
















Thanks my brain is messed up due to lack of sleep.
– Cliff
5 mins ago




Thanks my brain is messed up due to lack of sleep.
– Cliff
5 mins ago


















 

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