$int_9^16 f(x)dx = 64$, what is value of $int_9^16 f(x^2)xdx$

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Confused on the final step and want to double check my work.



Say $int_9^16 f(x)dx = 64$, what is value of $int_9^16 f(x^2)xdx$



I used u substitution and set $u = x^2$, then $fracdu2 = xdx$



Changed the bounds since $u=x^2$, take the square root of each.



Then $frac12int_3^4 f(u)du =$ ?




definite-integrals substitution




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  • 2




    no way of knowing. On the other hand, you could do $int_3^4 f(x^2)x dx$
    – Will Jagy
    Jul 26 at 0:42






  • 1




    When you integrate you will need change the bounds on your integral to accommodate your substitution...
    – Mason
    Jul 26 at 0:43














up vote
1
down vote

favorite












Confused on the final step and want to double check my work.



Say $int_9^16 f(x)dx = 64$, what is value of $int_9^16 f(x^2)xdx$



I used u substitution and set $u = x^2$, then $fracdu2 = xdx$



Changed the bounds since $u=x^2$, take the square root of each.



Then $frac12int_3^4 f(u)du =$ ?




definite-integrals substitution




share|cite|improve this question

















  • 2




    no way of knowing. On the other hand, you could do $int_3^4 f(x^2)x dx$
    – Will Jagy
    Jul 26 at 0:42






  • 1




    When you integrate you will need change the bounds on your integral to accommodate your substitution...
    – Mason
    Jul 26 at 0:43












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Confused on the final step and want to double check my work.



Say $int_9^16 f(x)dx = 64$, what is value of $int_9^16 f(x^2)xdx$



I used u substitution and set $u = x^2$, then $fracdu2 = xdx$



Changed the bounds since $u=x^2$, take the square root of each.



Then $frac12int_3^4 f(u)du =$ ?




definite-integrals substitution




share|cite|improve this question













Confused on the final step and want to double check my work.



Say $int_9^16 f(x)dx = 64$, what is value of $int_9^16 f(x^2)xdx$



I used u substitution and set $u = x^2$, then $fracdu2 = xdx$



Changed the bounds since $u=x^2$, take the square root of each.



Then $frac12int_3^4 f(u)du =$ ?





definite-integrals substitution





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 26 at 0:50
























asked Jul 26 at 0:32









Barto

62




62







  • 2




    no way of knowing. On the other hand, you could do $int_3^4 f(x^2)x dx$
    – Will Jagy
    Jul 26 at 0:42






  • 1




    When you integrate you will need change the bounds on your integral to accommodate your substitution...
    – Mason
    Jul 26 at 0:43












  • 2




    no way of knowing. On the other hand, you could do $int_3^4 f(x^2)x dx$
    – Will Jagy
    Jul 26 at 0:42






  • 1




    When you integrate you will need change the bounds on your integral to accommodate your substitution...
    – Mason
    Jul 26 at 0:43







2




2




no way of knowing. On the other hand, you could do $int_3^4 f(x^2)x dx$
– Will Jagy
Jul 26 at 0:42




no way of knowing. On the other hand, you could do $int_3^4 f(x^2)x dx$
– Will Jagy
Jul 26 at 0:42




1




1




When you integrate you will need change the bounds on your integral to accommodate your substitution...
– Mason
Jul 26 at 0:43




When you integrate you will need change the bounds on your integral to accommodate your substitution...
– Mason
Jul 26 at 0:43










1 Answer
1






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3
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No, you have not changed your limits of integration from $x$ to $u$.



Notice that if $$x=9, u=x^2=81$$



We simply do not have enough information to find $$ int_9^16 f(x^2)xdx$$






share|cite|improve this answer





















  • Ok, this makes sense to me now. Thank you.
    – Barto
    Jul 26 at 0:58










  • Thanks for your attention.
    – Mohammad Riazi-Kermani
    Jul 26 at 0:59










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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote













No, you have not changed your limits of integration from $x$ to $u$.



Notice that if $$x=9, u=x^2=81$$



We simply do not have enough information to find $$ int_9^16 f(x^2)xdx$$






share|cite|improve this answer





















  • Ok, this makes sense to me now. Thank you.
    – Barto
    Jul 26 at 0:58










  • Thanks for your attention.
    – Mohammad Riazi-Kermani
    Jul 26 at 0:59














up vote
3
down vote













No, you have not changed your limits of integration from $x$ to $u$.



Notice that if $$x=9, u=x^2=81$$



We simply do not have enough information to find $$ int_9^16 f(x^2)xdx$$






share|cite|improve this answer





















  • Ok, this makes sense to me now. Thank you.
    – Barto
    Jul 26 at 0:58










  • Thanks for your attention.
    – Mohammad Riazi-Kermani
    Jul 26 at 0:59












up vote
3
down vote










up vote
3
down vote









No, you have not changed your limits of integration from $x$ to $u$.



Notice that if $$x=9, u=x^2=81$$



We simply do not have enough information to find $$ int_9^16 f(x^2)xdx$$






share|cite|improve this answer













No, you have not changed your limits of integration from $x$ to $u$.



Notice that if $$x=9, u=x^2=81$$



We simply do not have enough information to find $$ int_9^16 f(x^2)xdx$$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 26 at 0:47









Mohammad Riazi-Kermani

27.3k41851




27.3k41851











  • Ok, this makes sense to me now. Thank you.
    – Barto
    Jul 26 at 0:58










  • Thanks for your attention.
    – Mohammad Riazi-Kermani
    Jul 26 at 0:59
















  • Ok, this makes sense to me now. Thank you.
    – Barto
    Jul 26 at 0:58










  • Thanks for your attention.
    – Mohammad Riazi-Kermani
    Jul 26 at 0:59















Ok, this makes sense to me now. Thank you.
– Barto
Jul 26 at 0:58




Ok, this makes sense to me now. Thank you.
– Barto
Jul 26 at 0:58












Thanks for your attention.
– Mohammad Riazi-Kermani
Jul 26 at 0:59




Thanks for your attention.
– Mohammad Riazi-Kermani
Jul 26 at 0:59












 

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