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$\lim _ { x \rightarrow 0 } \frac { \int _ { a x } ^ { b x } \left[ \int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s \right] d t } { \cos x - 1 }$

$L = \lim _ { x \rightarrow 0 } \frac { 1 } { \cos x - 1 } \left[ \int _ { a x } ^ { b x } \left( \int _ { c t } ^ { k t } e ^ { - s ^ { 2 } } d s \right) d t \right]$

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?

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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. $\endgroup$ Jan 30, 2019 at 8:44

2 Answers 2

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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 $(b^2-a^2)(c-k)$.

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  • $\begingroup$ thank , piece of cake :) $\endgroup$ Jan 30, 2019 at 9:50
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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 }$

$= \underset { x \rightarrow 0 } { \lim } \frac{\frac{d}{dx}[G(bx)-G(ax)-H(bx)+H(ax)]}{- sin x}$

$= \underset { x \rightarrow 0 } { \lim }\frac{b\cdot G\prime (bx)-a\cdot G\prime (ax)-b\cdot H\prime (bx)+a\cdot H\prime (ax)}{- \sin x}$

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 $ F\prime ( 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, $L = - b k b + c b ^ { 2 } - c a ^ { 2 } + k a ^ { 2 }$

$= b ^ { 2 } ( c - k ) + a ^ { 2 } ( k - c )$

$= ( k - c ) \left( a ^ { 2 } - b ^ { 2 } \right)$

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