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Definite integral giving negative value as a result?
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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 4 hours ago
Eevee Trainer
9,92731740
asked 4 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 4 hours ago
Eevee Trainer
9,92731740
asked 4 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
4 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
4 hours ago
$begingroup$
Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
$endgroup$
– weno
4 hours ago
5
$begingroup$
You flipped the interval's endpoints. $-2<-1$
$endgroup$
– mr_e_man
4 hours ago
$begingroup$
And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
$endgroup$
– Lubin
4 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 4 hours ago
Eevee Trainer
9,92731740
asked 4 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 4 hours ago
Eevee Trainer
9,92731740
asked 4 hours ago
wenoweno
39611
edited 4 hours ago
Eevee Trainer
9,92731740
asked 4 hours ago
wenoweno
39611
edited 4 hours ago
Eevee Trainer
9,92731740
edited 4 hours ago
Eevee Trainer
9,92731740
edited 4 hours ago
Eevee Trainer
9,92731740
9,92731740
asked 4 hours ago
wenoweno
39611
asked 4 hours ago
wenoweno
39611
asked 4 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
4 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
4 hours ago
$begingroup$
Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
$endgroup$
– weno
4 hours ago
5
$begingroup$
You flipped the interval's endpoints. $-2<-1$
$endgroup$
– mr_e_man
4 hours ago
$begingroup$
And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
$endgroup$
– Lubin
4 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
4 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
4 hours ago
$begingroup$
Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
$endgroup$
– weno
4 hours ago
5
$begingroup$
You flipped the interval's endpoints. $-2<-1$
$endgroup$
– mr_e_man
4 hours ago
$begingroup$
And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
$endgroup$
– Lubin
4 hours ago
2
2
$begingroup$
How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
$endgroup$
– Eevee Trainer
4 hours ago
$begingroup$
How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
$endgroup$
– Eevee Trainer
4 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
4 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
4 hours ago
$begingroup$
Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
$endgroup$
– weno
4 hours ago
$begingroup$
Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
$endgroup$
– weno
4 hours ago
5
5
$begingroup$
You flipped the interval's endpoints. $-2<-1$
$endgroup$
– mr_e_man
4 hours ago
$begingroup$
You flipped the interval's endpoints. $-2<-1$
$endgroup$
– mr_e_man
4 hours ago
$begingroup$
And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
$endgroup$
– Lubin
4 hours ago
$begingroup$
And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
$endgroup$
– Lubin
4 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 4 hours ago
Eevee TrainerEevee Trainer
9,92731740
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
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oldest
votes
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 4 hours ago
Eevee TrainerEevee Trainer
9,92731740
$endgroup$
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 4 hours ago
Eevee TrainerEevee Trainer
9,92731740
$endgroup$
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 4 hours ago
Eevee TrainerEevee Trainer
9,92731740
$endgroup$
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 4 hours ago
Eevee TrainerEevee Trainer
9,92731740
answered 4 hours ago
Eevee TrainerEevee Trainer
9,92731740
answered 4 hours ago
Eevee TrainerEevee Trainer
9,92731740
answered 4 hours ago
Eevee TrainerEevee Trainer
9,92731740
9,92731740
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
4 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
4 hours ago
$begingroup$
Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
$endgroup$
– weno
4 hours ago
5
$begingroup$
You flipped the interval's endpoints. $-2<-1$
$endgroup$
– mr_e_man
4 hours ago
$begingroup$
And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
$endgroup$
– Lubin
4 hours ago