Leibniz integration rule applied twice
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0
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I want to apply the Leibniz integration rule twice to this integral:
$$
frac{textrm d^2}{textrm d^2 x}int_a^x g(s)(x-s)ds
$$
The first application gives:
$$
frac{textrm d}{textrm d x}int_a^x g(s)(x-s)ds = int_a^xfrac{partial}{partial x}left(g(s)(x-s)right)ds = int_a^x g(s)ds
$$
But the next one troubles me:
$$
frac{textrm d}{textrm d x}int_a^x g(s)ds = int_a^xfrac{partial}{partial x}g(s)ds =
$$
The result should be
$$
g(x)
$$
But I cannot see why $g(x)$ is the result.
Any ideas?
calculus integration derivatives
edited Nov 27 at 6:24
qbert
22k32459
asked Nov 27 at 6:17
WolfgangP
1625
add a comment |
up vote
0
down vote
favorite
I want to apply the Leibniz integration rule twice to this integral:
$$
frac{textrm d^2}{textrm d^2 x}int_a^x g(s)(x-s)ds
$$
The first application gives:
$$
frac{textrm d}{textrm d x}int_a^x g(s)(x-s)ds = int_a^xfrac{partial}{partial x}left(g(s)(x-s)right)ds = int_a^x g(s)ds
$$
But the next one troubles me:
$$
frac{textrm d}{textrm d x}int_a^x g(s)ds = int_a^xfrac{partial}{partial x}g(s)ds =
$$
The result should be
$$
g(x)
$$
But I cannot see why $g(x)$ is the result.
Any ideas?
calculus integration derivatives
edited Nov 27 at 6:24
qbert
22k32459
asked Nov 27 at 6:17
WolfgangP
1625
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to apply the Leibniz integration rule twice to this integral:
$$
frac{textrm d^2}{textrm d^2 x}int_a^x g(s)(x-s)ds
$$
The first application gives:
$$
frac{textrm d}{textrm d x}int_a^x g(s)(x-s)ds = int_a^xfrac{partial}{partial x}left(g(s)(x-s)right)ds = int_a^x g(s)ds
$$
But the next one troubles me:
$$
frac{textrm d}{textrm d x}int_a^x g(s)ds = int_a^xfrac{partial}{partial x}g(s)ds =
$$
The result should be
$$
g(x)
$$
But I cannot see why $g(x)$ is the result.
Any ideas?
calculus integration derivatives
edited Nov 27 at 6:24
qbert
22k32459
asked Nov 27 at 6:17
WolfgangP
1625
I want to apply the Leibniz integration rule twice to this integral:
$$
frac{textrm d^2}{textrm d^2 x}int_a^x g(s)(x-s)ds
$$
The first application gives:
$$
frac{textrm d}{textrm d x}int_a^x g(s)(x-s)ds = int_a^xfrac{partial}{partial x}left(g(s)(x-s)right)ds = int_a^x g(s)ds
$$
But the next one troubles me:
$$
frac{textrm d}{textrm d x}int_a^x g(s)ds = int_a^xfrac{partial}{partial x}g(s)ds =
$$
The result should be
$$
g(x)
$$
But I cannot see why $g(x)$ is the result.
Any ideas?
calculus integration derivatives
calculus integration derivatives
edited Nov 27 at 6:24
qbert
22k32459
asked Nov 27 at 6:17
WolfgangP
1625
edited Nov 27 at 6:24
qbert
22k32459
asked Nov 27 at 6:17
WolfgangP
1625
edited Nov 27 at 6:24
qbert
22k32459
edited Nov 27 at 6:24
qbert
22k32459
edited Nov 27 at 6:24
qbert
22k32459
22k32459
asked Nov 27 at 6:17
WolfgangP
1625
asked Nov 27 at 6:17
WolfgangP
1625
asked Nov 27 at 6:17
WolfgangP
1625
1625
add a comment |
add a comment |
1 Answer
1
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up vote
2
down vote
accepted
What you wrote is not correct:
$$
frac{d}{dx}int_a^x g(s) mathrm dsneint_a^x frac{d}{dx}g(s)mathrm ds
$$
The result follows from the regular fundamental theorem of calculus
$$
frac{d}{dx}int_a^xg(s)mathrm ds=g(x)
$$
You may have gotten lucky in the first term and not been clear on how this rule works. In general,
$$
frac{d}{dx}int_{a(x)}^{b(x)}f(t,x)mathrm dt=f(b(x),x)b'(x)-f(a(x),x)a'(x)+int_{a(x)}^{b(x)}f_x(t,x)mathrm dt
$$
So in the first differentiation, you have
$$
frac{d}{dx}int_a^x g(s)(x-s)mathrm ds=g(x)(x-x)+int_a^x g(s)mathrm ds=
int_a^x g(s)mathrm ds
$$
answered Nov 27 at 6:23
qbert
22k32459
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
up vote
2
down vote
accepted
What you wrote is not correct:
$$
frac{d}{dx}int_a^x g(s) mathrm dsneint_a^x frac{d}{dx}g(s)mathrm ds
$$
The result follows from the regular fundamental theorem of calculus
$$
frac{d}{dx}int_a^xg(s)mathrm ds=g(x)
$$
You may have gotten lucky in the first term and not been clear on how this rule works. In general,
$$
frac{d}{dx}int_{a(x)}^{b(x)}f(t,x)mathrm dt=f(b(x),x)b'(x)-f(a(x),x)a'(x)+int_{a(x)}^{b(x)}f_x(t,x)mathrm dt
$$
So in the first differentiation, you have
$$
frac{d}{dx}int_a^x g(s)(x-s)mathrm ds=g(x)(x-x)+int_a^x g(s)mathrm ds=
int_a^x g(s)mathrm ds
$$
answered Nov 27 at 6:23
qbert
22k32459
add a comment |
up vote
2
down vote
accepted
What you wrote is not correct:
$$
frac{d}{dx}int_a^x g(s) mathrm dsneint_a^x frac{d}{dx}g(s)mathrm ds
$$
The result follows from the regular fundamental theorem of calculus
$$
frac{d}{dx}int_a^xg(s)mathrm ds=g(x)
$$
You may have gotten lucky in the first term and not been clear on how this rule works. In general,
$$
frac{d}{dx}int_{a(x)}^{b(x)}f(t,x)mathrm dt=f(b(x),x)b'(x)-f(a(x),x)a'(x)+int_{a(x)}^{b(x)}f_x(t,x)mathrm dt
$$
So in the first differentiation, you have
$$
frac{d}{dx}int_a^x g(s)(x-s)mathrm ds=g(x)(x-x)+int_a^x g(s)mathrm ds=
int_a^x g(s)mathrm ds
$$
answered Nov 27 at 6:23
qbert
22k32459
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
What you wrote is not correct:
$$
frac{d}{dx}int_a^x g(s) mathrm dsneint_a^x frac{d}{dx}g(s)mathrm ds
$$
The result follows from the regular fundamental theorem of calculus
$$
frac{d}{dx}int_a^xg(s)mathrm ds=g(x)
$$
You may have gotten lucky in the first term and not been clear on how this rule works. In general,
$$
frac{d}{dx}int_{a(x)}^{b(x)}f(t,x)mathrm dt=f(b(x),x)b'(x)-f(a(x),x)a'(x)+int_{a(x)}^{b(x)}f_x(t,x)mathrm dt
$$
So in the first differentiation, you have
$$
frac{d}{dx}int_a^x g(s)(x-s)mathrm ds=g(x)(x-x)+int_a^x g(s)mathrm ds=
int_a^x g(s)mathrm ds
$$
answered Nov 27 at 6:23
qbert
22k32459
What you wrote is not correct:
$$
frac{d}{dx}int_a^x g(s) mathrm dsneint_a^x frac{d}{dx}g(s)mathrm ds
$$
The result follows from the regular fundamental theorem of calculus
$$
frac{d}{dx}int_a^xg(s)mathrm ds=g(x)
$$
You may have gotten lucky in the first term and not been clear on how this rule works. In general,
$$
frac{d}{dx}int_{a(x)}^{b(x)}f(t,x)mathrm dt=f(b(x),x)b'(x)-f(a(x),x)a'(x)+int_{a(x)}^{b(x)}f_x(t,x)mathrm dt
$$
So in the first differentiation, you have
$$
frac{d}{dx}int_a^x g(s)(x-s)mathrm ds=g(x)(x-x)+int_a^x g(s)mathrm ds=
int_a^x g(s)mathrm ds
$$
answered Nov 27 at 6:23
qbert
22k32459
answered Nov 27 at 6:23
qbert
22k32459
answered Nov 27 at 6:23
qbert
22k32459
answered Nov 27 at 6:23
qbert
22k32459
22k32459
add a comment |
add a comment |
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