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Change of extremes of integration (integral function)
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If I have a certain $$F(x)=int_a(x)^b(x) f(t)dt$$
I know that $F'(x)=f(b(x))cdot b'(x) + f(a(x))cdot a'(x)$.
Now, my question is that: can I write $F$ as an integral function? So something like $$F(x)=int_p^x F'(t)dt=int_p^xf(b(t))cdot b'(t) + f(a(t))cdot a'(t)dt$$with $pin mathbbR$.
If it's possible, how do I find $p$?
integration
asked Jul 14 at 13:34
Besh00
827
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1
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If I have a certain $$F(x)=int_a(x)^b(x) f(t)dt$$
I know that $F'(x)=f(b(x))cdot b'(x) + f(a(x))cdot a'(x)$.
Now, my question is that: can I write $F$ as an integral function? So something like $$F(x)=int_p^x F'(t)dt=int_p^xf(b(t))cdot b'(t) + f(a(t))cdot a'(t)dt$$with $pin mathbbR$.
If it's possible, how do I find $p$?
integration
asked Jul 14 at 13:34
Besh00
827
1
It should be $f(b(x))cdot b’(x)-f(a(x))cdot a’(x)$
– Teh Rod
Jul 14 at 14:54
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If I have a certain $$F(x)=int_a(x)^b(x) f(t)dt$$
I know that $F'(x)=f(b(x))cdot b'(x) + f(a(x))cdot a'(x)$.
Now, my question is that: can I write $F$ as an integral function? So something like $$F(x)=int_p^x F'(t)dt=int_p^xf(b(t))cdot b'(t) + f(a(t))cdot a'(t)dt$$with $pin mathbbR$.
If it's possible, how do I find $p$?
integration
asked Jul 14 at 13:34
Besh00
827
If I have a certain $$F(x)=int_a(x)^b(x) f(t)dt$$
I know that $F'(x)=f(b(x))cdot b'(x) + f(a(x))cdot a'(x)$.
Now, my question is that: can I write $F$ as an integral function? So something like $$F(x)=int_p^x F'(t)dt=int_p^xf(b(t))cdot b'(t) + f(a(t))cdot a'(t)dt$$with $pin mathbbR$.
If it's possible, how do I find $p$?
integration
asked Jul 14 at 13:34
Besh00
827
asked Jul 14 at 13:34
Besh00
827
asked Jul 14 at 13:34
Besh00
827
asked Jul 14 at 13:34
Besh00
827
827
1
It should be $f(b(x))cdot b’(x)-f(a(x))cdot a’(x)$
– Teh Rod
Jul 14 at 14:54
add a comment |Â
1
It should be $f(b(x))cdot b’(x)-f(a(x))cdot a’(x)$
– Teh Rod
Jul 14 at 14:54
1
1
It should be $f(b(x))cdot b’(x)-f(a(x))cdot a’(x)$
– Teh Rod
Jul 14 at 14:54
It should be $f(b(x))cdot b’(x)-f(a(x))cdot a’(x)$
– Teh Rod
Jul 14 at 14:54
add a comment |Â
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1
It should be $f(b(x))cdot b’(x)-f(a(x))cdot a’(x)$
– Teh Rod
Jul 14 at 14:54