Mixing
Integrating positive function on interval from -1 to 1 but result is negative
$begingroup$
I have the following function:
$$int_{-1}^1 frac1 {x^4} dx$$
and my result seems to be:
$$frac {-2}3$$
Why is my result negative? My function is always positive, does it have to do with the fact that since the interval extends from -1 to 1 but the function doesn't go there that it somehow drastically affects the total integral?
calculus integration
asked Jan 18 at 4:27
Emma PascoeEmma Pascoe
161
$endgroup$
add a comment |
$begingroup$
I have the following function:
$$int_{-1}^1 frac1 {x^4} dx$$
and my result seems to be:
$$frac {-2}3$$
Why is my result negative? My function is always positive, does it have to do with the fact that since the interval extends from -1 to 1 but the function doesn't go there that it somehow drastically affects the total integral?
calculus integration
asked Jan 18 at 4:27
Emma PascoeEmma Pascoe
161
$endgroup$
$begingroup$
Welcome to MSE. Note that $frac{1}{x^4}$ is not defined at $x = 0$. Edit: On rereading your question, I believe you already knew this with your statement "the function doesn't go there".
$endgroup$
– John Omielan
Jan 18 at 4:29
$begingroup$
The integrand is not defined at zero.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 4:30
$begingroup$
There is a fatal discontinuity at $0$, so the FTC doesn't apply.
$endgroup$
– Randall
Jan 18 at 4:35
add a comment |
$begingroup$
I have the following function:
$$int_{-1}^1 frac1 {x^4} dx$$
and my result seems to be:
$$frac {-2}3$$
Why is my result negative? My function is always positive, does it have to do with the fact that since the interval extends from -1 to 1 but the function doesn't go there that it somehow drastically affects the total integral?
calculus integration
asked Jan 18 at 4:27
Emma PascoeEmma Pascoe
161
$endgroup$
I have the following function:
$$int_{-1}^1 frac1 {x^4} dx$$
and my result seems to be:
$$frac {-2}3$$
Why is my result negative? My function is always positive, does it have to do with the fact that since the interval extends from -1 to 1 but the function doesn't go there that it somehow drastically affects the total integral?
calculus integration
calculus integration
asked Jan 18 at 4:27
Emma PascoeEmma Pascoe
161
asked Jan 18 at 4:27
Emma PascoeEmma Pascoe
161
asked Jan 18 at 4:27
Emma PascoeEmma Pascoe
161
asked Jan 18 at 4:27
Emma PascoeEmma Pascoe
161
asked Jan 18 at 4:27
Emma PascoeEmma Pascoe
161
161
$begingroup$
Welcome to MSE. Note that $frac{1}{x^4}$ is not defined at $x = 0$. Edit: On rereading your question, I believe you already knew this with your statement "the function doesn't go there".
$endgroup$
– John Omielan
Jan 18 at 4:29
$begingroup$
The integrand is not defined at zero.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 4:30
$begingroup$
There is a fatal discontinuity at $0$, so the FTC doesn't apply.
$endgroup$
– Randall
Jan 18 at 4:35
add a comment |
$begingroup$
Welcome to MSE. Note that $frac{1}{x^4}$ is not defined at $x = 0$. Edit: On rereading your question, I believe you already knew this with your statement "the function doesn't go there".
$endgroup$
– John Omielan
Jan 18 at 4:29
$begingroup$
The integrand is not defined at zero.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 4:30
$begingroup$
There is a fatal discontinuity at $0$, so the FTC doesn't apply.
$endgroup$
– Randall
Jan 18 at 4:35
$begingroup$
Welcome to MSE. Note that $frac{1}{x^4}$ is not defined at $x = 0$. Edit: On rereading your question, I believe you already knew this with your statement "the function doesn't go there".
$endgroup$
– John Omielan
Jan 18 at 4:29
$begingroup$
Welcome to MSE. Note that $frac{1}{x^4}$ is not defined at $x = 0$. Edit: On rereading your question, I believe you already knew this with your statement "the function doesn't go there".
$endgroup$
– John Omielan
Jan 18 at 4:29
$begingroup$
The integrand is not defined at zero.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 4:30
$begingroup$
The integrand is not defined at zero.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 4:30
$begingroup$
There is a fatal discontinuity at $0$, so the FTC doesn't apply.
$endgroup$
– Randall
Jan 18 at 4:35
$begingroup$
There is a fatal discontinuity at $0$, so the FTC doesn't apply.
$endgroup$
– Randall
Jan 18 at 4:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since
$dfrac1{x^4}$
is not defined at
$x=0$,
one definition of
$int_{-1}^1 frac1 {x^4} dx
$
is
$lim_{c to 0^+} (int_{-1}^{-c} frac1 {x^4} dx+int_{c}^1 frac1 {x^4} dx)
$.
We have
$int dfrac{dx}{x^4}
=int x^{-4} dx
=dfrac{x^{-3}}{-3}
=-dfrac{1}{3x^3}
$
so
$begin{array}\
int_{-1}^{-c} frac1 {x^4} dx
&=-dfrac{1}{3x^3}|_{-1}^{-c}\
&=-(dfrac{1}{3(-c)^3}-dfrac{1}{3(-1)^3})\
&=-(-dfrac{1}{3c^3}+dfrac{1}{3})\
&=dfrac{1}{3c^3}-dfrac{1}{3}\
end{array}
$
and
$begin{array}\
int_{c}^{1} frac1 {x^4} dx
&=-dfrac{1}{3x^3}|_{c}^{1}\
&=-(dfrac{1}{3}-dfrac{1}{3c^3})\
&=dfrac{1}{3c^3}-dfrac{1}{3}\
end{array}
$
Adding these we get
$dfrac{2}{3c^3}-dfrac{2}{3}
$,
and the
$dfrac{2}{3c^3}
$
overwhelms the
$-dfrac{2}{3}
$.
answered Jan 18 at 4:47
marty cohenmarty cohen
74k549128
$endgroup$
1
$begingroup$
Why must the upper bound for the first integral be $-c$? Can't it be $c$ as well?
$endgroup$
– Aniruddh Venkatesan
Jan 18 at 4:50
1
$begingroup$
A good answer. One small thing to note is that as $frac{1}{x^4}$ is an even function, once you got the value of the first partial integral, you would know it's the exact same as the second partial integral.
$endgroup$
– John Omielan
Jan 18 at 4:51
1
$begingroup$
The interval of integration can't include zero because $1/x^4$ is not defined there.
$endgroup$
– marty cohen
Jan 18 at 8:20
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since
$dfrac1{x^4}$
is not defined at
$x=0$,
one definition of
$int_{-1}^1 frac1 {x^4} dx
$
is
$lim_{c to 0^+} (int_{-1}^{-c} frac1 {x^4} dx+int_{c}^1 frac1 {x^4} dx)
$.
We have
$int dfrac{dx}{x^4}
=int x^{-4} dx
=dfrac{x^{-3}}{-3}
=-dfrac{1}{3x^3}
$
so
$begin{array}\
int_{-1}^{-c} frac1 {x^4} dx
&=-dfrac{1}{3x^3}|_{-1}^{-c}\
&=-(dfrac{1}{3(-c)^3}-dfrac{1}{3(-1)^3})\
&=-(-dfrac{1}{3c^3}+dfrac{1}{3})\
&=dfrac{1}{3c^3}-dfrac{1}{3}\
end{array}
$
and
$begin{array}\
int_{c}^{1} frac1 {x^4} dx
&=-dfrac{1}{3x^3}|_{c}^{1}\
&=-(dfrac{1}{3}-dfrac{1}{3c^3})\
&=dfrac{1}{3c^3}-dfrac{1}{3}\
end{array}
$
Adding these we get
$dfrac{2}{3c^3}-dfrac{2}{3}
$,
and the
$dfrac{2}{3c^3}
$
overwhelms the
$-dfrac{2}{3}
$.
answered Jan 18 at 4:47
marty cohenmarty cohen
74k549128
$endgroup$
1
$begingroup$
Why must the upper bound for the first integral be $-c$? Can't it be $c$ as well?
$endgroup$
– Aniruddh Venkatesan
Jan 18 at 4:50
1
$begingroup$
A good answer. One small thing to note is that as $frac{1}{x^4}$ is an even function, once you got the value of the first partial integral, you would know it's the exact same as the second partial integral.
$endgroup$
– John Omielan
Jan 18 at 4:51
1
$begingroup$
The interval of integration can't include zero because $1/x^4$ is not defined there.
$endgroup$
– marty cohen
Jan 18 at 8:20
add a comment |
$begingroup$
Since
$dfrac1{x^4}$
is not defined at
$x=0$,
one definition of
$int_{-1}^1 frac1 {x^4} dx
$
is
$lim_{c to 0^+} (int_{-1}^{-c} frac1 {x^4} dx+int_{c}^1 frac1 {x^4} dx)
$.
We have
$int dfrac{dx}{x^4}
=int x^{-4} dx
=dfrac{x^{-3}}{-3}
=-dfrac{1}{3x^3}
$
so
$begin{array}\
int_{-1}^{-c} frac1 {x^4} dx
&=-dfrac{1}{3x^3}|_{-1}^{-c}\
&=-(dfrac{1}{3(-c)^3}-dfrac{1}{3(-1)^3})\
&=-(-dfrac{1}{3c^3}+dfrac{1}{3})\
&=dfrac{1}{3c^3}-dfrac{1}{3}\
end{array}
$
and
$begin{array}\
int_{c}^{1} frac1 {x^4} dx
&=-dfrac{1}{3x^3}|_{c}^{1}\
&=-(dfrac{1}{3}-dfrac{1}{3c^3})\
&=dfrac{1}{3c^3}-dfrac{1}{3}\
end{array}
$
Adding these we get
$dfrac{2}{3c^3}-dfrac{2}{3}
$,
and the
$dfrac{2}{3c^3}
$
overwhelms the
$-dfrac{2}{3}
$.
answered Jan 18 at 4:47
marty cohenmarty cohen
74k549128
$endgroup$
1
$begingroup$
Why must the upper bound for the first integral be $-c$? Can't it be $c$ as well?
$endgroup$
– Aniruddh Venkatesan
Jan 18 at 4:50
1
$begingroup$
A good answer. One small thing to note is that as $frac{1}{x^4}$ is an even function, once you got the value of the first partial integral, you would know it's the exact same as the second partial integral.
$endgroup$
– John Omielan
Jan 18 at 4:51
1
$begingroup$
The interval of integration can't include zero because $1/x^4$ is not defined there.
$endgroup$
– marty cohen
Jan 18 at 8:20
add a comment |
$begingroup$
Since
$dfrac1{x^4}$
is not defined at
$x=0$,
one definition of
$int_{-1}^1 frac1 {x^4} dx
$
is
$lim_{c to 0^+} (int_{-1}^{-c} frac1 {x^4} dx+int_{c}^1 frac1 {x^4} dx)
$.
We have
$int dfrac{dx}{x^4}
=int x^{-4} dx
=dfrac{x^{-3}}{-3}
=-dfrac{1}{3x^3}
$
so
$begin{array}\
int_{-1}^{-c} frac1 {x^4} dx
&=-dfrac{1}{3x^3}|_{-1}^{-c}\
&=-(dfrac{1}{3(-c)^3}-dfrac{1}{3(-1)^3})\
&=-(-dfrac{1}{3c^3}+dfrac{1}{3})\
&=dfrac{1}{3c^3}-dfrac{1}{3}\
end{array}
$
and
$begin{array}\
int_{c}^{1} frac1 {x^4} dx
&=-dfrac{1}{3x^3}|_{c}^{1}\
&=-(dfrac{1}{3}-dfrac{1}{3c^3})\
&=dfrac{1}{3c^3}-dfrac{1}{3}\
end{array}
$
Adding these we get
$dfrac{2}{3c^3}-dfrac{2}{3}
$,
and the
$dfrac{2}{3c^3}
$
overwhelms the
$-dfrac{2}{3}
$.
answered Jan 18 at 4:47
marty cohenmarty cohen
74k549128
$endgroup$
Since
$dfrac1{x^4}$
is not defined at
$x=0$,
one definition of
$int_{-1}^1 frac1 {x^4} dx
$
is
$lim_{c to 0^+} (int_{-1}^{-c} frac1 {x^4} dx+int_{c}^1 frac1 {x^4} dx)
$.
We have
$int dfrac{dx}{x^4}
=int x^{-4} dx
=dfrac{x^{-3}}{-3}
=-dfrac{1}{3x^3}
$
so
$begin{array}\
int_{-1}^{-c} frac1 {x^4} dx
&=-dfrac{1}{3x^3}|_{-1}^{-c}\
&=-(dfrac{1}{3(-c)^3}-dfrac{1}{3(-1)^3})\
&=-(-dfrac{1}{3c^3}+dfrac{1}{3})\
&=dfrac{1}{3c^3}-dfrac{1}{3}\
end{array}
$
and
$begin{array}\
int_{c}^{1} frac1 {x^4} dx
&=-dfrac{1}{3x^3}|_{c}^{1}\
&=-(dfrac{1}{3}-dfrac{1}{3c^3})\
&=dfrac{1}{3c^3}-dfrac{1}{3}\
end{array}
$
Adding these we get
$dfrac{2}{3c^3}-dfrac{2}{3}
$,
and the
$dfrac{2}{3c^3}
$
overwhelms the
$-dfrac{2}{3}
$.
answered Jan 18 at 4:47
marty cohenmarty cohen
74k549128
answered Jan 18 at 4:47
marty cohenmarty cohen
74k549128
answered Jan 18 at 4:47
marty cohenmarty cohen
74k549128
answered Jan 18 at 4:47
marty cohenmarty cohen
74k549128
74k549128
1
$begingroup$
Why must the upper bound for the first integral be $-c$? Can't it be $c$ as well?
$endgroup$
– Aniruddh Venkatesan
Jan 18 at 4:50
1
$begingroup$
A good answer. One small thing to note is that as $frac{1}{x^4}$ is an even function, once you got the value of the first partial integral, you would know it's the exact same as the second partial integral.
$endgroup$
– John Omielan
Jan 18 at 4:51
1
$begingroup$
The interval of integration can't include zero because $1/x^4$ is not defined there.
$endgroup$
– marty cohen
Jan 18 at 8:20
add a comment |
1
$begingroup$
Why must the upper bound for the first integral be $-c$? Can't it be $c$ as well?
$endgroup$
– Aniruddh Venkatesan
Jan 18 at 4:50
1
$begingroup$
A good answer. One small thing to note is that as $frac{1}{x^4}$ is an even function, once you got the value of the first partial integral, you would know it's the exact same as the second partial integral.
$endgroup$
– John Omielan
Jan 18 at 4:51
1
$begingroup$
The interval of integration can't include zero because $1/x^4$ is not defined there.
$endgroup$
– marty cohen
Jan 18 at 8:20
1
1
$begingroup$
Why must the upper bound for the first integral be $-c$? Can't it be $c$ as well?
$endgroup$
– Aniruddh Venkatesan
Jan 18 at 4:50
$begingroup$
Why must the upper bound for the first integral be $-c$? Can't it be $c$ as well?
$endgroup$
– Aniruddh Venkatesan
Jan 18 at 4:50
1
1
$begingroup$
A good answer. One small thing to note is that as $frac{1}{x^4}$ is an even function, once you got the value of the first partial integral, you would know it's the exact same as the second partial integral.
$endgroup$
– John Omielan
Jan 18 at 4:51
$begingroup$
A good answer. One small thing to note is that as $frac{1}{x^4}$ is an even function, once you got the value of the first partial integral, you would know it's the exact same as the second partial integral.
$endgroup$
– John Omielan
Jan 18 at 4:51
1
1
$begingroup$
The interval of integration can't include zero because $1/x^4$ is not defined there.
$endgroup$
– marty cohen
Jan 18 at 8:20
$begingroup$
The interval of integration can't include zero because $1/x^4$ is not defined there.
$endgroup$
– marty cohen
Jan 18 at 8:20
add a comment |
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$begingroup$
Welcome to MSE. Note that $frac{1}{x^4}$ is not defined at $x = 0$. Edit: On rereading your question, I believe you already knew this with your statement "the function doesn't go there".
$endgroup$
– John Omielan
Jan 18 at 4:29
$begingroup$
The integrand is not defined at zero.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 4:30
$begingroup$
There is a fatal discontinuity at $0$, so the FTC doesn't apply.
$endgroup$
– Randall
Jan 18 at 4:35