Mixing
if a, b, c, k ∈ R , Compute the following limit
$begingroup$
$lim _ { x rightarrow 0 } frac { int _ { a x } ^ { b x } left[ int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s right] d t } { cos x - 1 }$
$L = lim _ { x rightarrow 0 } frac { 1 } { cos x - 1 } left[ int _ { a x } ^ { b x } left( int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s right) d t right]$
If we plugin limit, we will get 0/0 form.
Hence we can apply LHospitals rule.
diff denominator wrt x and diff numerator wrt t
$L = lim _ { x rightarrow 0 } frac { 1 } { - sin x } left[ int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s right] _ { a x } ^ { b x }$
plugin the limits
$L = lim _ { x rightarrow 0 } frac { 1 } { - sin x } left[ int _ { c ( b x ) } ^ { k ( b x ) } e ^ { - s ^ { 2 } } d s - int _ { c ( a x ) } ^ { k ( a x ) } e ^ { - s ^ { 2 } } d s right]$
$L = lim _ { x rightarrow 0 } frac { 1 } { - sin x } left[ int _ { c b x } ^ { k b x } e ^ { - s ^ { 2 } } d s - int _ { c a x } ^ { k a x } e ^ { - s ^ { 2 } } d s right]$
f we plugin limit, we will get 0/0 form.
Hence we can apply LHospitals rule again.
diff denominator wrt x and diff numerator wrt t
$L = lim _ { x rightarrow 0 } frac { 1 } { - cos x } left( left[ - 2 s e ^ { - s ^ { 2 } } right] _ { c b x } ^ { k b x } - left[ - 2 s e ^ { - s ^ { 2 } } right] _ { c a x } ^ { k a x } right)$
multiply the -ve sign
$L = lim _ { x rightarrow 0 } frac { 1 } { cos x } left( left[ - 2 s e ^ { - s ^ { 2 } } right] _ { operatorname { cax } } ^ { k a x } - left[ - 2 s e ^ { - s ^ { 2 } } right] _ { c b x } ^ { k b x } right)$
remove -ve signs and reverse limits
$L = lim _ { x rightarrow 0 } frac { 1 } { cos x } left( left[ 2 s e ^ { - s ^ { 2 } } right] _ { k a x } ^ { operatorname { cax } } - left[ 2 s e ^ { - s ^ { 2 } } right] _ { k b x } ^ { c b x } right)$
Take 2 common
$L = lim _ { x rightarrow 0 } frac { 2 } { cos x } left( left[ s e ^ { - s ^ { 2 } } right] _ { k a x } ^ { operatorname { cax } } - left[ s e ^ { - s ^ { 2 } } right] _ { k b x } ^ { c b x } right)$
plugin the limits
$L = lim _ { x rightarrow 0 } frac { 2 } { cos x } left( left[ operatorname { caxe } ^ { - c ^ { 2 } a ^ { 2 } x ^ { 2 } } - k a x e ^ { - k ^ { 2 } a ^ { 2 } x ^ { 2 } } right] - left[ c b x e ^ { - c ^ { 2 } b ^ { 2 } x ^ { 2 } } - k b x e ^ { - k ^ { 2 } b ^ { 2 } x ^ { 2 } } right] right)$
plugin the limit for x
$L = frac { 2 } { cos 0 } ( [ 0 - 0 ] - [ 0 - 0 ] )$
$L = frac { 2 } { 1 } ( 0 ) = 0$
Is this correct?
definite-integrals indefinite-integrals
asked Jan 30 at 8:22
Tariro ManyikaTariro Manyika
619
$endgroup$
add a comment |
$begingroup$
$lim _ { x rightarrow 0 } frac { int _ { a x } ^ { b x } left[ int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s right] d t } { cos x - 1 }$
$L = lim _ { x rightarrow 0 } frac { 1 } { cos x - 1 } left[ int _ { a x } ^ { b x } left( int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s right) d t right]$
If we plugin limit, we will get 0/0 form.
Hence we can apply LHospitals rule.
diff denominator wrt x and diff numerator wrt t
$L = lim _ { x rightarrow 0 } frac { 1 } { - sin x } left[ int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s right] _ { a x } ^ { b x }$
plugin the limits
$L = lim _ { x rightarrow 0 } frac { 1 } { - sin x } left[ int _ { c ( b x ) } ^ { k ( b x ) } e ^ { - s ^ { 2 } } d s - int _ { c ( a x ) } ^ { k ( a x ) } e ^ { - s ^ { 2 } } d s right]$
$L = lim _ { x rightarrow 0 } frac { 1 } { - sin x } left[ int _ { c b x } ^ { k b x } e ^ { - s ^ { 2 } } d s - int _ { c a x } ^ { k a x } e ^ { - s ^ { 2 } } d s right]$
f we plugin limit, we will get 0/0 form.
Hence we can apply LHospitals rule again.
diff denominator wrt x and diff numerator wrt t
$L = lim _ { x rightarrow 0 } frac { 1 } { - cos x } left( left[ - 2 s e ^ { - s ^ { 2 } } right] _ { c b x } ^ { k b x } - left[ - 2 s e ^ { - s ^ { 2 } } right] _ { c a x } ^ { k a x } right)$
multiply the -ve sign
$L = lim _ { x rightarrow 0 } frac { 1 } { cos x } left( left[ - 2 s e ^ { - s ^ { 2 } } right] _ { operatorname { cax } } ^ { k a x } - left[ - 2 s e ^ { - s ^ { 2 } } right] _ { c b x } ^ { k b x } right)$
remove -ve signs and reverse limits
$L = lim _ { x rightarrow 0 } frac { 1 } { cos x } left( left[ 2 s e ^ { - s ^ { 2 } } right] _ { k a x } ^ { operatorname { cax } } - left[ 2 s e ^ { - s ^ { 2 } } right] _ { k b x } ^ { c b x } right)$
Take 2 common
$L = lim _ { x rightarrow 0 } frac { 2 } { cos x } left( left[ s e ^ { - s ^ { 2 } } right] _ { k a x } ^ { operatorname { cax } } - left[ s e ^ { - s ^ { 2 } } right] _ { k b x } ^ { c b x } right)$
plugin the limits
$L = lim _ { x rightarrow 0 } frac { 2 } { cos x } left( left[ operatorname { caxe } ^ { - c ^ { 2 } a ^ { 2 } x ^ { 2 } } - k a x e ^ { - k ^ { 2 } a ^ { 2 } x ^ { 2 } } right] - left[ c b x e ^ { - c ^ { 2 } b ^ { 2 } x ^ { 2 } } - k b x e ^ { - k ^ { 2 } b ^ { 2 } x ^ { 2 } } right] right)$
plugin the limit for x
$L = frac { 2 } { cos 0 } ( [ 0 - 0 ] - [ 0 - 0 ] )$
$L = frac { 2 } { 1 } ( 0 ) = 0$
Is this correct?
definite-integrals indefinite-integrals
asked Jan 30 at 8:22
Tariro ManyikaTariro Manyika
619
$endgroup$
1
$begingroup$
Are you supposed to know the result of $int e^{-s^2},ds$ ? If yes, you could make the problem much shorter.
$endgroup$
– Claude Leibovici
Jan 30 at 8:44
add a comment |
$begingroup$
$lim _ { x rightarrow 0 } frac { int _ { a x } ^ { b x } left[ int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s right] d t } { cos x - 1 }$
$L = lim _ { x rightarrow 0 } frac { 1 } { cos x - 1 } left[ int _ { a x } ^ { b x } left( int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s right) d t right]$
If we plugin limit, we will get 0/0 form.
Hence we can apply LHospitals rule.
diff denominator wrt x and diff numerator wrt t
$L = lim _ { x rightarrow 0 } frac { 1 } { - sin x } left[ int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s right] _ { a x } ^ { b x }$
plugin the limits
$L = lim _ { x rightarrow 0 } frac { 1 } { - sin x } left[ int _ { c ( b x ) } ^ { k ( b x ) } e ^ { - s ^ { 2 } } d s - int _ { c ( a x ) } ^ { k ( a x ) } e ^ { - s ^ { 2 } } d s right]$
$L = lim _ { x rightarrow 0 } frac { 1 } { - sin x } left[ int _ { c b x } ^ { k b x } e ^ { - s ^ { 2 } } d s - int _ { c a x } ^ { k a x } e ^ { - s ^ { 2 } } d s right]$
f we plugin limit, we will get 0/0 form.
Hence we can apply LHospitals rule again.
diff denominator wrt x and diff numerator wrt t
$L = lim _ { x rightarrow 0 } frac { 1 } { - cos x } left( left[ - 2 s e ^ { - s ^ { 2 } } right] _ { c b x } ^ { k b x } - left[ - 2 s e ^ { - s ^ { 2 } } right] _ { c a x } ^ { k a x } right)$
multiply the -ve sign
$L = lim _ { x rightarrow 0 } frac { 1 } { cos x } left( left[ - 2 s e ^ { - s ^ { 2 } } right] _ { operatorname { cax } } ^ { k a x } - left[ - 2 s e ^ { - s ^ { 2 } } right] _ { c b x } ^ { k b x } right)$
remove -ve signs and reverse limits
$L = lim _ { x rightarrow 0 } frac { 1 } { cos x } left( left[ 2 s e ^ { - s ^ { 2 } } right] _ { k a x } ^ { operatorname { cax } } - left[ 2 s e ^ { - s ^ { 2 } } right] _ { k b x } ^ { c b x } right)$
Take 2 common
$L = lim _ { x rightarrow 0 } frac { 2 } { cos x } left( left[ s e ^ { - s ^ { 2 } } right] _ { k a x } ^ { operatorname { cax } } - left[ s e ^ { - s ^ { 2 } } right] _ { k b x } ^ { c b x } right)$
plugin the limits
$L = lim _ { x rightarrow 0 } frac { 2 } { cos x } left( left[ operatorname { caxe } ^ { - c ^ { 2 } a ^ { 2 } x ^ { 2 } } - k a x e ^ { - k ^ { 2 } a ^ { 2 } x ^ { 2 } } right] - left[ c b x e ^ { - c ^ { 2 } b ^ { 2 } x ^ { 2 } } - k b x e ^ { - k ^ { 2 } b ^ { 2 } x ^ { 2 } } right] right)$
plugin the limit for x
$L = frac { 2 } { cos 0 } ( [ 0 - 0 ] - [ 0 - 0 ] )$
$L = frac { 2 } { 1 } ( 0 ) = 0$
Is this correct?
definite-integrals indefinite-integrals
asked Jan 30 at 8:22
Tariro ManyikaTariro Manyika
619
$endgroup$
$lim _ { x rightarrow 0 } frac { int _ { a x } ^ { b x } left[ int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s right] d t } { cos x - 1 }$
$L = lim _ { x rightarrow 0 } frac { 1 } { cos x - 1 } left[ int _ { a x } ^ { b x } left( int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s right) d t right]$
If we plugin limit, we will get 0/0 form.
Hence we can apply LHospitals rule.
diff denominator wrt x and diff numerator wrt t
$L = lim _ { x rightarrow 0 } frac { 1 } { - sin x } left[ int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s right] _ { a x } ^ { b x }$
plugin the limits
$L = lim _ { x rightarrow 0 } frac { 1 } { - sin x } left[ int _ { c ( b x ) } ^ { k ( b x ) } e ^ { - s ^ { 2 } } d s - int _ { c ( a x ) } ^ { k ( a x ) } e ^ { - s ^ { 2 } } d s right]$
$L = lim _ { x rightarrow 0 } frac { 1 } { - sin x } left[ int _ { c b x } ^ { k b x } e ^ { - s ^ { 2 } } d s - int _ { c a x } ^ { k a x } e ^ { - s ^ { 2 } } d s right]$
f we plugin limit, we will get 0/0 form.
Hence we can apply LHospitals rule again.
diff denominator wrt x and diff numerator wrt t
$L = lim _ { x rightarrow 0 } frac { 1 } { - cos x } left( left[ - 2 s e ^ { - s ^ { 2 } } right] _ { c b x } ^ { k b x } - left[ - 2 s e ^ { - s ^ { 2 } } right] _ { c a x } ^ { k a x } right)$
multiply the -ve sign
$L = lim _ { x rightarrow 0 } frac { 1 } { cos x } left( left[ - 2 s e ^ { - s ^ { 2 } } right] _ { operatorname { cax } } ^ { k a x } - left[ - 2 s e ^ { - s ^ { 2 } } right] _ { c b x } ^ { k b x } right)$
remove -ve signs and reverse limits
$L = lim _ { x rightarrow 0 } frac { 1 } { cos x } left( left[ 2 s e ^ { - s ^ { 2 } } right] _ { k a x } ^ { operatorname { cax } } - left[ 2 s e ^ { - s ^ { 2 } } right] _ { k b x } ^ { c b x } right)$
Take 2 common
$L = lim _ { x rightarrow 0 } frac { 2 } { cos x } left( left[ s e ^ { - s ^ { 2 } } right] _ { k a x } ^ { operatorname { cax } } - left[ s e ^ { - s ^ { 2 } } right] _ { k b x } ^ { c b x } right)$
plugin the limits
$L = lim _ { x rightarrow 0 } frac { 2 } { cos x } left( left[ operatorname { caxe } ^ { - c ^ { 2 } a ^ { 2 } x ^ { 2 } } - k a x e ^ { - k ^ { 2 } a ^ { 2 } x ^ { 2 } } right] - left[ c b x e ^ { - c ^ { 2 } b ^ { 2 } x ^ { 2 } } - k b x e ^ { - k ^ { 2 } b ^ { 2 } x ^ { 2 } } right] right)$
plugin the limit for x
$L = frac { 2 } { cos 0 } ( [ 0 - 0 ] - [ 0 - 0 ] )$
$L = frac { 2 } { 1 } ( 0 ) = 0$
Is this correct?
definite-integrals indefinite-integrals
definite-integrals indefinite-integrals
asked Jan 30 at 8:22
Tariro ManyikaTariro Manyika
619
asked Jan 30 at 8:22
Tariro ManyikaTariro Manyika
619
edited Jan 30 at 8:31
Tariro Manyika
asked Jan 30 at 8:22
Tariro ManyikaTariro Manyika
619
asked Jan 30 at 8:22
Tariro ManyikaTariro Manyika
619
asked Jan 30 at 8:22
Tariro ManyikaTariro Manyika
619
619
1
$begingroup$
Are you supposed to know the result of $int e^{-s^2},ds$ ? If yes, you could make the problem much shorter.
$endgroup$
– Claude Leibovici
Jan 30 at 8:44
add a comment |
1
$begingroup$
Are you supposed to know the result of $int e^{-s^2},ds$ ? If yes, you could make the problem much shorter.
$endgroup$
– Claude Leibovici
Jan 30 at 8:44
1
1
$begingroup$
Are you supposed to know the result of $int e^{-s^2},ds$ ? If yes, you could make the problem much shorter.
$endgroup$
– Claude Leibovici
Jan 30 at 8:44
$begingroup$
Are you supposed to know the result of $int e^{-s^2},ds$ ? If yes, you could make the problem much shorter.
$endgroup$
– Claude Leibovici
Jan 30 at 8:44
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No... Keep in mind that
$$
frac{d}{dx} int_{alpha(x)}^{beta(x)} f(s) ds = beta'(x) f(beta(x))-alpha'(x) f(alpha(x))
$$
so you see that you are missing some multiplicative constants relative to the derivatives of the integration limits. The correct answer is $(b^2-a^2)(c-k)$.
answered Jan 30 at 8:37
PierreCarrePierreCarre
1,692212
$endgroup$
$begingroup$
thank , piece of cake :)
$endgroup$
– Tariro Manyika
Jan 30 at 9:50
add a comment |
$begingroup$
Let $int e ^ { - s ^ { 2 } } d s = F ( s ) Rightarrow frac { d } { d s } ( F ( s ) ) = e ^ { - s ^ { 2 } }$
Then $int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s = F left. ( s ) right| _ { c t } ^ { k t }$
$= F ( k t ) - F ( c t )$
So numerator becomes :-
$int _ { a x } ^ { b x } left[ int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s right] d t = int _ { a x } ^ { b x } [ F ( k t ) - F ( c t ) ] d t$
$= int _ { a x } ^ { b x } F ( k t ) d t - int _ { a x } ^ { b x } F ( c t ) d t$
So we need to evaluate the limit :-
$L = lim _ { x rightarrow 0 } frac { int _ { a x } ^ { b x } F ( k t ) d t - int _ { a x } ^ { b x } F ( c t ) d t } { cos x - 1 }$
This is a $frac{0}{0}$ form. So, we can use L'Hospital's rule as:-
$ L = lim _ { x rightarrow 0 } frac { frac { d } { d x } left[ int _ { a x } ^ { b x } F ( k t ) d t - int _ { a x } ^ { b x } F ( c t ) d t right] } { frac { d } { d x } ( cos x - 1 ) }$
Now, let $int F ( k t ) d t = G ( t )$ and $int F ( c t ) d t = H ( t )$
such that: $frac { d } { d t } ( G ( t ) ) = F ( k t ) $ and $ frac { d } { d t } ( H ( t ) ) = F ( c t )$
So, $ L = lim _ { x rightarrow 0 } frac { frac { d } { d x } left[ G left. ( t ) right| _ { a x } ^ { b x } - H left. ( t ) right| _ { a x } ^ { b x } right] } { - sin x }$
$= underset { x rightarrow 0 } { lim } frac{frac{d}{dx}[G(bx)-G(ax)-H(bx)+H(ax)]}{- sin x}$
$= underset { x rightarrow 0 } { lim }frac{bcdot Gprime (bx)-acdot Gprime (ax)-bcdot Hprime (bx)+acdot Hprime (ax)}{- sin x}$
But as $G ^ { prime } ( t ) = F ( k t )$ and $H ^ { prime } ( t ) = F ( c t ) , so$:-
$L = lim _ { x rightarrow 0 } frac { b cdot F ( k x b ) - a cdot F ( k x a ) - b cdot F ( c b x ) + a cdot F ( c a x ) } { - sin x }$
$ L = lim _ { x rightarrow 0 } frac { b [ F ( k x b ) - F ( c x b ) ] + a [ F ( c a x ) - F ( k x a ) ] } { - sin x }$
This is still $frac{0}{0}$ form. So use L'Hopital's once more to get
$ L = lim _ { x rightarrow 0 } frac {frac{d}{dx}[ b [ F ( k x b ) - F ( c x b ) ] + a [ F ( c a x ) - F ( k x a ) ]] } { frac{d}{dx}(- sin x) }$
$ L = lim _ { x rightarrow 0 } frac { b frac{d}{dx} [ F ( k x b ) - F ( c x b ) ] + a frac{d}{dx} [ F ( c a x ) - F ( k x a ) ] } { frac{d}{dx}(- cos x) }$
As $ Fprime ( s ) = e ^ { - s ^ { 2 } }$ so :-
$frac { d } { d x } ( F ( k x b ) ) = k b cdot F ^ { prime } ( kxb ) = k b e ^ { - ( k a b ) ^ { 2 } }$
$frac { d } { d x } ( F ( c x b ) ) = c b cdot F ^ { prime } ( cxb ) = c b e ^ { - ( c x b ) ^ { 2 } }$
$frac { d } { d x } ( F ( c x a ) ) = c a cdot F ^ { prime } ( cxa ) = c a e ^ { - ( c x a ) ^ { 2 } }$
$frac { d } { d x } ( F ( k x a ) ) = k a cdot F ^ { prime } ( kxa ) = k a e ^ { - ( k x a ) ^ { 2 } }$
and also, $lim _ { x rightarrow 0 } ( - cos x ) = - 1$ So,
$L = lim _ { x rightarrow 0 } - [ b ( k b e ^ { - ( k x b ) ^ { 2 } } - c b e ^ { - (cxb)^{2}} + a ( c a e ^ { - (cax )^{2} } - k a e ^ { - ( k x a ) ^ { 2 } } )$
$= - [ b ( k b - c b ) + a ( c a - k a ) ]$
Because $lim _ { x rightarrow 0 } e ^ { - x ^ { 2 } } = e ^ { 0 } = 1$
So, $L = - b k b + c b ^ { 2 } - c a ^ { 2 } + k a ^ { 2 }$
$= b ^ { 2 } ( c - k ) + a ^ { 2 } ( k - c )$
$= ( k - c ) left( a ^ { 2 } - b ^ { 2 } right)$
answered Jan 31 at 15:39
Tariro ManyikaTariro Manyika
619
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
No... Keep in mind that
$$
frac{d}{dx} int_{alpha(x)}^{beta(x)} f(s) ds = beta'(x) f(beta(x))-alpha'(x) f(alpha(x))
$$
so you see that you are missing some multiplicative constants relative to the derivatives of the integration limits. The correct answer is $(b^2-a^2)(c-k)$.
answered Jan 30 at 8:37
PierreCarrePierreCarre
1,692212
$endgroup$
$begingroup$
thank , piece of cake :)
$endgroup$
– Tariro Manyika
Jan 30 at 9:50
add a comment |
$begingroup$
No... Keep in mind that
$$
frac{d}{dx} int_{alpha(x)}^{beta(x)} f(s) ds = beta'(x) f(beta(x))-alpha'(x) f(alpha(x))
$$
so you see that you are missing some multiplicative constants relative to the derivatives of the integration limits. The correct answer is $(b^2-a^2)(c-k)$.
answered Jan 30 at 8:37
PierreCarrePierreCarre
1,692212
$endgroup$
$begingroup$
thank , piece of cake :)
$endgroup$
– Tariro Manyika
Jan 30 at 9:50
add a comment |
$begingroup$
No... Keep in mind that
$$
frac{d}{dx} int_{alpha(x)}^{beta(x)} f(s) ds = beta'(x) f(beta(x))-alpha'(x) f(alpha(x))
$$
so you see that you are missing some multiplicative constants relative to the derivatives of the integration limits. The correct answer is $(b^2-a^2)(c-k)$.
answered Jan 30 at 8:37
PierreCarrePierreCarre
1,692212
$endgroup$
No... Keep in mind that
$$
frac{d}{dx} int_{alpha(x)}^{beta(x)} f(s) ds = beta'(x) f(beta(x))-alpha'(x) f(alpha(x))
$$
so you see that you are missing some multiplicative constants relative to the derivatives of the integration limits. The correct answer is $(b^2-a^2)(c-k)$.
answered Jan 30 at 8:37
PierreCarrePierreCarre
1,692212
answered Jan 30 at 8:37
PierreCarrePierreCarre
1,692212
answered Jan 30 at 8:37
PierreCarrePierreCarre
1,692212
answered Jan 30 at 8:37
PierreCarrePierreCarre
1,692212
1,692212
$begingroup$
thank , piece of cake :)
$endgroup$
– Tariro Manyika
Jan 30 at 9:50
add a comment |
$begingroup$
thank , piece of cake :)
$endgroup$
– Tariro Manyika
Jan 30 at 9:50
$begingroup$
thank , piece of cake :)
$endgroup$
– Tariro Manyika
Jan 30 at 9:50
$begingroup$
thank , piece of cake :)
$endgroup$
– Tariro Manyika
Jan 30 at 9:50
add a comment |
$begingroup$
Let $int e ^ { - s ^ { 2 } } d s = F ( s ) Rightarrow frac { d } { d s } ( F ( s ) ) = e ^ { - s ^ { 2 } }$
Then $int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s = F left. ( s ) right| _ { c t } ^ { k t }$
$= F ( k t ) - F ( c t )$
So numerator becomes :-
$int _ { a x } ^ { b x } left[ int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s right] d t = int _ { a x } ^ { b x } [ F ( k t ) - F ( c t ) ] d t$
$= int _ { a x } ^ { b x } F ( k t ) d t - int _ { a x } ^ { b x } F ( c t ) d t$
So we need to evaluate the limit :-
$L = lim _ { x rightarrow 0 } frac { int _ { a x } ^ { b x } F ( k t ) d t - int _ { a x } ^ { b x } F ( c t ) d t } { cos x - 1 }$
This is a $frac{0}{0}$ form. So, we can use L'Hospital's rule as:-
$ L = lim _ { x rightarrow 0 } frac { frac { d } { d x } left[ int _ { a x } ^ { b x } F ( k t ) d t - int _ { a x } ^ { b x } F ( c t ) d t right] } { frac { d } { d x } ( cos x - 1 ) }$
Now, let $int F ( k t ) d t = G ( t )$ and $int F ( c t ) d t = H ( t )$
such that: $frac { d } { d t } ( G ( t ) ) = F ( k t ) $ and $ frac { d } { d t } ( H ( t ) ) = F ( c t )$
So, $ L = lim _ { x rightarrow 0 } frac { frac { d } { d x } left[ G left. ( t ) right| _ { a x } ^ { b x } - H left. ( t ) right| _ { a x } ^ { b x } right] } { - sin x }$
$= underset { x rightarrow 0 } { lim } frac{frac{d}{dx}[G(bx)-G(ax)-H(bx)+H(ax)]}{- sin x}$
$= underset { x rightarrow 0 } { lim }frac{bcdot Gprime (bx)-acdot Gprime (ax)-bcdot Hprime (bx)+acdot Hprime (ax)}{- sin x}$
But as $G ^ { prime } ( t ) = F ( k t )$ and $H ^ { prime } ( t ) = F ( c t ) , so$:-
$L = lim _ { x rightarrow 0 } frac { b cdot F ( k x b ) - a cdot F ( k x a ) - b cdot F ( c b x ) + a cdot F ( c a x ) } { - sin x }$
$ L = lim _ { x rightarrow 0 } frac { b [ F ( k x b ) - F ( c x b ) ] + a [ F ( c a x ) - F ( k x a ) ] } { - sin x }$
This is still $frac{0}{0}$ form. So use L'Hopital's once more to get
$ L = lim _ { x rightarrow 0 } frac {frac{d}{dx}[ b [ F ( k x b ) - F ( c x b ) ] + a [ F ( c a x ) - F ( k x a ) ]] } { frac{d}{dx}(- sin x) }$
$ L = lim _ { x rightarrow 0 } frac { b frac{d}{dx} [ F ( k x b ) - F ( c x b ) ] + a frac{d}{dx} [ F ( c a x ) - F ( k x a ) ] } { frac{d}{dx}(- cos x) }$
As $ Fprime ( s ) = e ^ { - s ^ { 2 } }$ so :-
$frac { d } { d x } ( F ( k x b ) ) = k b cdot F ^ { prime } ( kxb ) = k b e ^ { - ( k a b ) ^ { 2 } }$
$frac { d } { d x } ( F ( c x b ) ) = c b cdot F ^ { prime } ( cxb ) = c b e ^ { - ( c x b ) ^ { 2 } }$
$frac { d } { d x } ( F ( c x a ) ) = c a cdot F ^ { prime } ( cxa ) = c a e ^ { - ( c x a ) ^ { 2 } }$
$frac { d } { d x } ( F ( k x a ) ) = k a cdot F ^ { prime } ( kxa ) = k a e ^ { - ( k x a ) ^ { 2 } }$
and also, $lim _ { x rightarrow 0 } ( - cos x ) = - 1$ So,
$L = lim _ { x rightarrow 0 } - [ b ( k b e ^ { - ( k x b ) ^ { 2 } } - c b e ^ { - (cxb)^{2}} + a ( c a e ^ { - (cax )^{2} } - k a e ^ { - ( k x a ) ^ { 2 } } )$
$= - [ b ( k b - c b ) + a ( c a - k a ) ]$
Because $lim _ { x rightarrow 0 } e ^ { - x ^ { 2 } } = e ^ { 0 } = 1$
So, $L = - b k b + c b ^ { 2 } - c a ^ { 2 } + k a ^ { 2 }$
$= b ^ { 2 } ( c - k ) + a ^ { 2 } ( k - c )$
$= ( k - c ) left( a ^ { 2 } - b ^ { 2 } right)$
answered Jan 31 at 15:39
Tariro ManyikaTariro Manyika
619
$endgroup$
add a comment |
$begingroup$
Let $int e ^ { - s ^ { 2 } } d s = F ( s ) Rightarrow frac { d } { d s } ( F ( s ) ) = e ^ { - s ^ { 2 } }$
Then $int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s = F left. ( s ) right| _ { c t } ^ { k t }$
$= F ( k t ) - F ( c t )$
So numerator becomes :-
$int _ { a x } ^ { b x } left[ int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s right] d t = int _ { a x } ^ { b x } [ F ( k t ) - F ( c t ) ] d t$
$= int _ { a x } ^ { b x } F ( k t ) d t - int _ { a x } ^ { b x } F ( c t ) d t$
So we need to evaluate the limit :-
$L = lim _ { x rightarrow 0 } frac { int _ { a x } ^ { b x } F ( k t ) d t - int _ { a x } ^ { b x } F ( c t ) d t } { cos x - 1 }$
This is a $frac{0}{0}$ form. So, we can use L'Hospital's rule as:-
$ L = lim _ { x rightarrow 0 } frac { frac { d } { d x } left[ int _ { a x } ^ { b x } F ( k t ) d t - int _ { a x } ^ { b x } F ( c t ) d t right] } { frac { d } { d x } ( cos x - 1 ) }$
Now, let $int F ( k t ) d t = G ( t )$ and $int F ( c t ) d t = H ( t )$
such that: $frac { d } { d t } ( G ( t ) ) = F ( k t ) $ and $ frac { d } { d t } ( H ( t ) ) = F ( c t )$
So, $ L = lim _ { x rightarrow 0 } frac { frac { d } { d x } left[ G left. ( t ) right| _ { a x } ^ { b x } - H left. ( t ) right| _ { a x } ^ { b x } right] } { - sin x }$
$= underset { x rightarrow 0 } { lim } frac{frac{d}{dx}[G(bx)-G(ax)-H(bx)+H(ax)]}{- sin x}$
$= underset { x rightarrow 0 } { lim }frac{bcdot Gprime (bx)-acdot Gprime (ax)-bcdot Hprime (bx)+acdot Hprime (ax)}{- sin x}$
But as $G ^ { prime } ( t ) = F ( k t )$ and $H ^ { prime } ( t ) = F ( c t ) , so$:-
$L = lim _ { x rightarrow 0 } frac { b cdot F ( k x b ) - a cdot F ( k x a ) - b cdot F ( c b x ) + a cdot F ( c a x ) } { - sin x }$
$ L = lim _ { x rightarrow 0 } frac { b [ F ( k x b ) - F ( c x b ) ] + a [ F ( c a x ) - F ( k x a ) ] } { - sin x }$
This is still $frac{0}{0}$ form. So use L'Hopital's once more to get
$ L = lim _ { x rightarrow 0 } frac {frac{d}{dx}[ b [ F ( k x b ) - F ( c x b ) ] + a [ F ( c a x ) - F ( k x a ) ]] } { frac{d}{dx}(- sin x) }$
$ L = lim _ { x rightarrow 0 } frac { b frac{d}{dx} [ F ( k x b ) - F ( c x b ) ] + a frac{d}{dx} [ F ( c a x ) - F ( k x a ) ] } { frac{d}{dx}(- cos x) }$
As $ Fprime ( s ) = e ^ { - s ^ { 2 } }$ so :-
$frac { d } { d x } ( F ( k x b ) ) = k b cdot F ^ { prime } ( kxb ) = k b e ^ { - ( k a b ) ^ { 2 } }$
$frac { d } { d x } ( F ( c x b ) ) = c b cdot F ^ { prime } ( cxb ) = c b e ^ { - ( c x b ) ^ { 2 } }$
$frac { d } { d x } ( F ( c x a ) ) = c a cdot F ^ { prime } ( cxa ) = c a e ^ { - ( c x a ) ^ { 2 } }$
$frac { d } { d x } ( F ( k x a ) ) = k a cdot F ^ { prime } ( kxa ) = k a e ^ { - ( k x a ) ^ { 2 } }$
and also, $lim _ { x rightarrow 0 } ( - cos x ) = - 1$ So,
$L = lim _ { x rightarrow 0 } - [ b ( k b e ^ { - ( k x b ) ^ { 2 } } - c b e ^ { - (cxb)^{2}} + a ( c a e ^ { - (cax )^{2} } - k a e ^ { - ( k x a ) ^ { 2 } } )$
$= - [ b ( k b - c b ) + a ( c a - k a ) ]$
Because $lim _ { x rightarrow 0 } e ^ { - x ^ { 2 } } = e ^ { 0 } = 1$
So, $L = - b k b + c b ^ { 2 } - c a ^ { 2 } + k a ^ { 2 }$
$= b ^ { 2 } ( c - k ) + a ^ { 2 } ( k - c )$
$= ( k - c ) left( a ^ { 2 } - b ^ { 2 } right)$
answered Jan 31 at 15:39
Tariro ManyikaTariro Manyika
619
$endgroup$
add a comment |
$begingroup$
Let $int e ^ { - s ^ { 2 } } d s = F ( s ) Rightarrow frac { d } { d s } ( F ( s ) ) = e ^ { - s ^ { 2 } }$
Then $int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s = F left. ( s ) right| _ { c t } ^ { k t }$
$= F ( k t ) - F ( c t )$
So numerator becomes :-
$int _ { a x } ^ { b x } left[ int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s right] d t = int _ { a x } ^ { b x } [ F ( k t ) - F ( c t ) ] d t$
$= int _ { a x } ^ { b x } F ( k t ) d t - int _ { a x } ^ { b x } F ( c t ) d t$
So we need to evaluate the limit :-
$L = lim _ { x rightarrow 0 } frac { int _ { a x } ^ { b x } F ( k t ) d t - int _ { a x } ^ { b x } F ( c t ) d t } { cos x - 1 }$
This is a $frac{0}{0}$ form. So, we can use L'Hospital's rule as:-
$ L = lim _ { x rightarrow 0 } frac { frac { d } { d x } left[ int _ { a x } ^ { b x } F ( k t ) d t - int _ { a x } ^ { b x } F ( c t ) d t right] } { frac { d } { d x } ( cos x - 1 ) }$
Now, let $int F ( k t ) d t = G ( t )$ and $int F ( c t ) d t = H ( t )$
such that: $frac { d } { d t } ( G ( t ) ) = F ( k t ) $ and $ frac { d } { d t } ( H ( t ) ) = F ( c t )$
So, $ L = lim _ { x rightarrow 0 } frac { frac { d } { d x } left[ G left. ( t ) right| _ { a x } ^ { b x } - H left. ( t ) right| _ { a x } ^ { b x } right] } { - sin x }$
$= underset { x rightarrow 0 } { lim } frac{frac{d}{dx}[G(bx)-G(ax)-H(bx)+H(ax)]}{- sin x}$
$= underset { x rightarrow 0 } { lim }frac{bcdot Gprime (bx)-acdot Gprime (ax)-bcdot Hprime (bx)+acdot Hprime (ax)}{- sin x}$
But as $G ^ { prime } ( t ) = F ( k t )$ and $H ^ { prime } ( t ) = F ( c t ) , so$:-
$L = lim _ { x rightarrow 0 } frac { b cdot F ( k x b ) - a cdot F ( k x a ) - b cdot F ( c b x ) + a cdot F ( c a x ) } { - sin x }$
$ L = lim _ { x rightarrow 0 } frac { b [ F ( k x b ) - F ( c x b ) ] + a [ F ( c a x ) - F ( k x a ) ] } { - sin x }$
This is still $frac{0}{0}$ form. So use L'Hopital's once more to get
$ L = lim _ { x rightarrow 0 } frac {frac{d}{dx}[ b [ F ( k x b ) - F ( c x b ) ] + a [ F ( c a x ) - F ( k x a ) ]] } { frac{d}{dx}(- sin x) }$
$ L = lim _ { x rightarrow 0 } frac { b frac{d}{dx} [ F ( k x b ) - F ( c x b ) ] + a frac{d}{dx} [ F ( c a x ) - F ( k x a ) ] } { frac{d}{dx}(- cos x) }$
As $ Fprime ( s ) = e ^ { - s ^ { 2 } }$ so :-
$frac { d } { d x } ( F ( k x b ) ) = k b cdot F ^ { prime } ( kxb ) = k b e ^ { - ( k a b ) ^ { 2 } }$
$frac { d } { d x } ( F ( c x b ) ) = c b cdot F ^ { prime } ( cxb ) = c b e ^ { - ( c x b ) ^ { 2 } }$
$frac { d } { d x } ( F ( c x a ) ) = c a cdot F ^ { prime } ( cxa ) = c a e ^ { - ( c x a ) ^ { 2 } }$
$frac { d } { d x } ( F ( k x a ) ) = k a cdot F ^ { prime } ( kxa ) = k a e ^ { - ( k x a ) ^ { 2 } }$
and also, $lim _ { x rightarrow 0 } ( - cos x ) = - 1$ So,
$L = lim _ { x rightarrow 0 } - [ b ( k b e ^ { - ( k x b ) ^ { 2 } } - c b e ^ { - (cxb)^{2}} + a ( c a e ^ { - (cax )^{2} } - k a e ^ { - ( k x a ) ^ { 2 } } )$
$= - [ b ( k b - c b ) + a ( c a - k a ) ]$
Because $lim _ { x rightarrow 0 } e ^ { - x ^ { 2 } } = e ^ { 0 } = 1$
So, $L = - b k b + c b ^ { 2 } - c a ^ { 2 } + k a ^ { 2 }$
$= b ^ { 2 } ( c - k ) + a ^ { 2 } ( k - c )$
$= ( k - c ) left( a ^ { 2 } - b ^ { 2 } right)$
answered Jan 31 at 15:39
Tariro ManyikaTariro Manyika
619
$endgroup$
Let $int e ^ { - s ^ { 2 } } d s = F ( s ) Rightarrow frac { d } { d s } ( F ( s ) ) = e ^ { - s ^ { 2 } }$
Then $int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s = F left. ( s ) right| _ { c t } ^ { k t }$
$= F ( k t ) - F ( c t )$
So numerator becomes :-
$int _ { a x } ^ { b x } left[ int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s right] d t = int _ { a x } ^ { b x } [ F ( k t ) - F ( c t ) ] d t$
$= int _ { a x } ^ { b x } F ( k t ) d t - int _ { a x } ^ { b x } F ( c t ) d t$
So we need to evaluate the limit :-
$L = lim _ { x rightarrow 0 } frac { int _ { a x } ^ { b x } F ( k t ) d t - int _ { a x } ^ { b x } F ( c t ) d t } { cos x - 1 }$
This is a $frac{0}{0}$ form. So, we can use L'Hospital's rule as:-
$ L = lim _ { x rightarrow 0 } frac { frac { d } { d x } left[ int _ { a x } ^ { b x } F ( k t ) d t - int _ { a x } ^ { b x } F ( c t ) d t right] } { frac { d } { d x } ( cos x - 1 ) }$
Now, let $int F ( k t ) d t = G ( t )$ and $int F ( c t ) d t = H ( t )$
such that: $frac { d } { d t } ( G ( t ) ) = F ( k t ) $ and $ frac { d } { d t } ( H ( t ) ) = F ( c t )$
So, $ L = lim _ { x rightarrow 0 } frac { frac { d } { d x } left[ G left. ( t ) right| _ { a x } ^ { b x } - H left. ( t ) right| _ { a x } ^ { b x } right] } { - sin x }$
$= underset { x rightarrow 0 } { lim } frac{frac{d}{dx}[G(bx)-G(ax)-H(bx)+H(ax)]}{- sin x}$
$= underset { x rightarrow 0 } { lim }frac{bcdot Gprime (bx)-acdot Gprime (ax)-bcdot Hprime (bx)+acdot Hprime (ax)}{- sin x}$
But as $G ^ { prime } ( t ) = F ( k t )$ and $H ^ { prime } ( t ) = F ( c t ) , so$:-
$L = lim _ { x rightarrow 0 } frac { b cdot F ( k x b ) - a cdot F ( k x a ) - b cdot F ( c b x ) + a cdot F ( c a x ) } { - sin x }$
$ L = lim _ { x rightarrow 0 } frac { b [ F ( k x b ) - F ( c x b ) ] + a [ F ( c a x ) - F ( k x a ) ] } { - sin x }$
This is still $frac{0}{0}$ form. So use L'Hopital's once more to get
$ L = lim _ { x rightarrow 0 } frac {frac{d}{dx}[ b [ F ( k x b ) - F ( c x b ) ] + a [ F ( c a x ) - F ( k x a ) ]] } { frac{d}{dx}(- sin x) }$
$ L = lim _ { x rightarrow 0 } frac { b frac{d}{dx} [ F ( k x b ) - F ( c x b ) ] + a frac{d}{dx} [ F ( c a x ) - F ( k x a ) ] } { frac{d}{dx}(- cos x) }$
As $ Fprime ( s ) = e ^ { - s ^ { 2 } }$ so :-
$frac { d } { d x } ( F ( k x b ) ) = k b cdot F ^ { prime } ( kxb ) = k b e ^ { - ( k a b ) ^ { 2 } }$
$frac { d } { d x } ( F ( c x b ) ) = c b cdot F ^ { prime } ( cxb ) = c b e ^ { - ( c x b ) ^ { 2 } }$
$frac { d } { d x } ( F ( c x a ) ) = c a cdot F ^ { prime } ( cxa ) = c a e ^ { - ( c x a ) ^ { 2 } }$
$frac { d } { d x } ( F ( k x a ) ) = k a cdot F ^ { prime } ( kxa ) = k a e ^ { - ( k x a ) ^ { 2 } }$
and also, $lim _ { x rightarrow 0 } ( - cos x ) = - 1$ So,
$L = lim _ { x rightarrow 0 } - [ b ( k b e ^ { - ( k x b ) ^ { 2 } } - c b e ^ { - (cxb)^{2}} + a ( c a e ^ { - (cax )^{2} } - k a e ^ { - ( k x a ) ^ { 2 } } )$
$= - [ b ( k b - c b ) + a ( c a - k a ) ]$
Because $lim _ { x rightarrow 0 } e ^ { - x ^ { 2 } } = e ^ { 0 } = 1$
So, $L = - b k b + c b ^ { 2 } - c a ^ { 2 } + k a ^ { 2 }$
$= b ^ { 2 } ( c - k ) + a ^ { 2 } ( k - c )$
$= ( k - c ) left( a ^ { 2 } - b ^ { 2 } right)$
answered Jan 31 at 15:39
Tariro ManyikaTariro Manyika
619
answered Jan 31 at 15:39
Tariro ManyikaTariro Manyika
619
answered Jan 31 at 15:39
Tariro ManyikaTariro Manyika
619
answered Jan 31 at 15:39
Tariro ManyikaTariro Manyika
619
619
add a comment |
add a comment |
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$begingroup$
Are you supposed to know the result of $int e^{-s^2},ds$ ? If yes, you could make the problem much shorter.
$endgroup$
– Claude Leibovici
Jan 30 at 8:44