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Definite integral giving negative value as a result?
Why do I get a negative value for this integral?Solving a definite integralReal integral giving a complex resultProgression from indefinite integral to definite integral - $int_0^2pifrac15-3cos x dx$Calculation of definite integralWithout calculating the integral decide if integral is positive or negative / which integral is bigger?Definite integral of absolute value function?Variable substitution in definite integralDefinite integral over singularityInner Product, Definite Integral
$begingroup$
I want to calculate definite integral
$$int_-2^-1 frac1x^2e^frac1x dx = Omega$$
$$int frac1x^2e^frac1x dx=-e^frac1x+C$$
so:
$$Omega = [-e^frac1-2]-[-e^frac1-1]=-frac1sqrte + frac1e$$
which is a negative value. I believe it should be positive.
What went wrong in the process?
calculus integration definite-integrals
edited 23 hours ago
Eevee Trainer
10k31740
asked 23 hours ago
wenoweno
39611
$endgroup$
|
show 2 more comments
$begingroup$
I want to calculate definite integral
$$int_-2^-1 frac1x^2e^frac1x dx = Omega$$
$$int frac1x^2e^frac1x dx=-e^frac1x+C$$
so:
$$Omega = [-e^frac1-2]-[-e^frac1-1]=-frac1sqrte + frac1e$$
which is a negative value. I believe it should be positive.
What went wrong in the process?
calculus integration definite-integrals
edited 23 hours ago
Eevee Trainer
10k31740
asked 23 hours ago
wenoweno
39611
$endgroup$
2
$begingroup$
How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
$endgroup$
– Eevee Trainer
23 hours ago
2
$begingroup$
Your antiderivative is completely incorrect: The derivative of $e^1/x^2$ is $e^1/x^2 / (-x^3)$. The red flag that you found is indeed a correct one, and shows that your answer cannot be right. This is a good thing to check.
$endgroup$
– T. Bongers
23 hours ago
$begingroup$
Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
$endgroup$
– weno
23 hours ago
5
$begingroup$
You flipped the interval's endpoints. $-2<-1$
$endgroup$
– mr_e_man
23 hours ago
$begingroup$
And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
$endgroup$
– Lubin
23 hours ago
|
show 2 more comments
$begingroup$
I want to calculate definite integral
$$int_-2^-1 frac1x^2e^frac1x dx = Omega$$
$$int frac1x^2e^frac1x dx=-e^frac1x+C$$
so:
$$Omega = [-e^frac1-2]-[-e^frac1-1]=-frac1sqrte + frac1e$$
which is a negative value. I believe it should be positive.
What went wrong in the process?
calculus integration definite-integrals
edited 23 hours ago
Eevee Trainer
10k31740
asked 23 hours ago
wenoweno
39611
$endgroup$
I want to calculate definite integral
$$int_-2^-1 frac1x^2e^frac1x dx = Omega$$
$$int frac1x^2e^frac1x dx=-e^frac1x+C$$
so:
$$Omega = [-e^frac1-2]-[-e^frac1-1]=-frac1sqrte + frac1e$$
which is a negative value. I believe it should be positive.
What went wrong in the process?
calculus integration definite-integrals
calculus integration definite-integrals
edited 23 hours ago
Eevee Trainer
10k31740
asked 23 hours ago
wenoweno
39611
edited 23 hours ago
Eevee Trainer
10k31740
asked 23 hours ago
wenoweno
39611
edited 23 hours ago
Eevee Trainer
10k31740
edited 23 hours ago
Eevee Trainer
10k31740
edited 23 hours ago
Eevee Trainer
10k31740
10k31740
asked 23 hours ago
wenoweno
39611
asked 23 hours ago
wenoweno
39611
asked 23 hours ago
wenoweno
39611
39611
2
$begingroup$
How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
$endgroup$
– Eevee Trainer
23 hours ago
2
$begingroup$
Your antiderivative is completely incorrect: The derivative of $e^1/x^2$ is $e^1/x^2 / (-x^3)$. The red flag that you found is indeed a correct one, and shows that your answer cannot be right. This is a good thing to check.
$endgroup$
– T. Bongers
23 hours ago
$begingroup$
Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
$endgroup$
– weno
23 hours ago
5
$begingroup$
You flipped the interval's endpoints. $-2<-1$
$endgroup$
– mr_e_man
23 hours ago
$begingroup$
And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
$endgroup$
– Lubin
23 hours ago
|
show 2 more comments
2
$begingroup$
How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
$endgroup$
– Eevee Trainer
23 hours ago
2
$begingroup$
Your antiderivative is completely incorrect: The derivative of $e^1/x^2$ is $e^1/x^2 / (-x^3)$. The red flag that you found is indeed a correct one, and shows that your answer cannot be right. This is a good thing to check.
$endgroup$
– T. Bongers
23 hours ago
$begingroup$
Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
$endgroup$
– weno
23 hours ago
5
$begingroup$
You flipped the interval's endpoints. $-2<-1$
$endgroup$
– mr_e_man
23 hours ago
$begingroup$
And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
$endgroup$
– Lubin
23 hours ago
2
2
$begingroup$
How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
$endgroup$
– Eevee Trainer
23 hours ago
$begingroup$
How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
$endgroup$
– Eevee Trainer
23 hours ago
2
2
$begingroup$
Your antiderivative is completely incorrect: The derivative of $e^1/x^2$ is $e^1/x^2 / (-x^3)$. The red flag that you found is indeed a correct one, and shows that your answer cannot be right. This is a good thing to check.
$endgroup$
– T. Bongers
23 hours ago
$begingroup$
Your antiderivative is completely incorrect: The derivative of $e^1/x^2$ is $e^1/x^2 / (-x^3)$. The red flag that you found is indeed a correct one, and shows that your answer cannot be right. This is a good thing to check.
$endgroup$
– T. Bongers
23 hours ago
$begingroup$
Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
$endgroup$
– weno
23 hours ago
$begingroup$
Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
$endgroup$
– weno
23 hours ago
5
5
$begingroup$
You flipped the interval's endpoints. $-2<-1$
$endgroup$
– mr_e_man
23 hours ago
$begingroup$
You flipped the interval's endpoints. $-2<-1$
$endgroup$
– mr_e_man
23 hours ago
$begingroup$
And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
$endgroup$
– Lubin
23 hours ago
$begingroup$
And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
$endgroup$
– Lubin
23 hours ago
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
What you effectively did was swap the order of evaluation for the fundamental theorem of calculus. Recall:
$$int_a^b f(x)dx = F(b) - F(a)$$
when the antiderivative of $f$ is $F$. You instead have $F(a) - F(b)$ ($a=-2,b=-1$) in this case. The end result is merely a sign error - you have precisely the negative of the answer which you should expect.
answered 23 hours ago
Eevee TrainerEevee Trainer
10k31740
$endgroup$
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
What you effectively did was swap the order of evaluation for the fundamental theorem of calculus. Recall:
$$int_a^b f(x)dx = F(b) - F(a)$$
when the antiderivative of $f$ is $F$. You instead have $F(a) - F(b)$ ($a=-2,b=-1$) in this case. The end result is merely a sign error - you have precisely the negative of the answer which you should expect.
answered 23 hours ago
Eevee TrainerEevee Trainer
10k31740
$endgroup$
add a comment |
$begingroup$
What you effectively did was swap the order of evaluation for the fundamental theorem of calculus. Recall:
$$int_a^b f(x)dx = F(b) - F(a)$$
when the antiderivative of $f$ is $F$. You instead have $F(a) - F(b)$ ($a=-2,b=-1$) in this case. The end result is merely a sign error - you have precisely the negative of the answer which you should expect.
answered 23 hours ago
Eevee TrainerEevee Trainer
10k31740
$endgroup$
add a comment |
$begingroup$
What you effectively did was swap the order of evaluation for the fundamental theorem of calculus. Recall:
$$int_a^b f(x)dx = F(b) - F(a)$$
when the antiderivative of $f$ is $F$. You instead have $F(a) - F(b)$ ($a=-2,b=-1$) in this case. The end result is merely a sign error - you have precisely the negative of the answer which you should expect.
answered 23 hours ago
Eevee TrainerEevee Trainer
10k31740
$endgroup$
What you effectively did was swap the order of evaluation for the fundamental theorem of calculus. Recall:
$$int_a^b f(x)dx = F(b) - F(a)$$
when the antiderivative of $f$ is $F$. You instead have $F(a) - F(b)$ ($a=-2,b=-1$) in this case. The end result is merely a sign error - you have precisely the negative of the answer which you should expect.
answered 23 hours ago
Eevee TrainerEevee Trainer
10k31740
answered 23 hours ago
Eevee TrainerEevee Trainer
10k31740
answered 23 hours ago
Eevee TrainerEevee Trainer
10k31740
answered 23 hours ago
Eevee TrainerEevee Trainer
10k31740
10k31740
add a comment |
add a comment |
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2
$begingroup$
How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
$endgroup$
– Eevee Trainer
23 hours ago
2
$begingroup$
Your antiderivative is completely incorrect: The derivative of $e^1/x^2$ is $e^1/x^2 / (-x^3)$. The red flag that you found is indeed a correct one, and shows that your answer cannot be right. This is a good thing to check.
$endgroup$
– T. Bongers
23 hours ago
$begingroup$
Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
$endgroup$
– weno
23 hours ago
5
$begingroup$
You flipped the interval's endpoints. $-2<-1$
$endgroup$
– mr_e_man
23 hours ago
$begingroup$
And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
$endgroup$
– Lubin
23 hours ago