Mixing
Area between the curves of $2cos(x)$ and $x/2$
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1
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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
asked 1 hour ago
Cliff
84
|
show 3 more comments
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
asked 1 hour ago
Cliff
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
|
show 3 more comments
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
asked 1 hour ago
Cliff
84
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
asked 1 hour ago
Cliff
84
asked 1 hour ago
Cliff
84
edited 46 mins ago
asked 1 hour ago
Cliff
84
asked 1 hour ago
Cliff
84
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
|
show 3 more comments
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
|
show 3 more comments
1 Answer
1
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.
answered 19 mins ago
Chris Custer
8,5692623
Thanks my brain is messed up due to lack of sleep.
– Cliff
5 mins ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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.
answered 19 mins ago
Chris Custer
8,5692623
Thanks my brain is messed up due to lack of sleep.
– Cliff
5 mins ago
add a comment |
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.
answered 19 mins ago
Chris Custer
8,5692623
Thanks my brain is messed up due to lack of sleep.
– Cliff
5 mins ago
add a comment |
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.
answered 19 mins ago
Chris Custer
8,5692623
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.
answered 19 mins ago
Chris Custer
8,5692623
answered 19 mins ago
Chris Custer
8,5692623
answered 19 mins ago
Chris Custer
8,5692623
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
add a comment |
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
add a comment |
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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