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@mathslover
mathslover
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It is easy , it just looks long. I will do that after lunch, ok?
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\[\Huge \lim_{x \rightarrow 0}( \frac{1}{x} \int\limits_{0}^{a} e^{\sin^{2}t}dt-\int\limits_{x+y}^{a} e^{\sin^{2}t}dt)\]
DLS
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Okay @mathslover and I know it is easy,I was trying it but i think the first term will cancel off because a and y both are constants maybe..i can be wrong,just drop a hint.
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sorry the term 1/x is outside everything
yrelhan4
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we can write the expression as.... (first term) + ∫ e^(sin^2 t dt) with limits ... a to x+y.. the sign changed on reversing the limits..
yrelhan4
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now combining the two limits.. 0 to a and a to x+y .. 0 to x+y ..
yrelhan4
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because the integrand is same..
DLS
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0 to a and 0 to x+y=0 to x+y
yrelhan4
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yep.
DLS
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cool :O
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\[\Huge \frac{1}{x} e^{\sin^{2}x+y}\]
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differenatiate and bingo :O
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e^sin^2x+y
yrelhan4
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why did you not do anything to 1/x ?
yrelhan4
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multiplication theorem.. hmm..
DLS
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1/x was outside the integral sorry :P
yrelhan4
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aur jab 0 put krenge.. to bhi to kuch aayega. vo kahan gyi? :P
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Wait.Lets start again :|
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\[\Huge \lim_{x \rightarrow 0} \frac{1}{x}(\int\limits_{y}^{a}e^{\sin^{2}t} dt-\int\limits_{x+y}^{a}e^{\sin^{2}t} dt)\]
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original question^
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\[\Large \lim_{x \rightarrow 0} \frac{1}{x}(\int\limits_{y}^{x+y}e^{\sin^{2}t} dt)\]
On changing the sign and revevrsing the limits and combining them
yrelhan4
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acha vo 'a' vala to 0 ho jayega.. a ki differentiation 0.. but hamne x+y ko differentiate bhi nhi kra.. :/
DLS
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Apply LH Rule
yrelhan4
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@shubhamsrg ?? :)
shubhamsrg
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kya hua? kya dikkat hai ? (:
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\[\Huge e^{\sin^{2}{x+y}}(1+\frac{dy}{dx})-e^{\sin^{2}y} \frac{dy}{dx}\]
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put dy/dx=0
yrelhan4
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why put dy/dx=0 ?
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\[\Huge e^{\sin^{2}x+y}\]
answer
DLS
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y is behaving as a constant,we assume y independent of x
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idk what i wrote please someone explain :P
yrelhan4
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ohhh.. achaaaa..
DLS
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If we are driving a car and chasing a thief on a bike then the rate of change of our speed won't affect the speed of the theif's bike
yrelhan4
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limit y se x+y ?? how??
DLS
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tune he to bataya?:/
yrelhan4
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i said 0 to x+y.. :/
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y->x+y hoga
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wo "A" hai 0 nahi :|
yrelhan4
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ohh. ok.
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:)
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but still didnt get it :|
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\[\LARGE e^{\sin^{2}x+y}\]
differentiating this^
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\[\LARGE e^{\sin^{2}x+y} \times \sin2x(+0)\]
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\[\LARGE e^{\sin^{2}y}\]
Differentiating the lower limit^
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\[\LARGE e^{\sin^{2}y}=>e^{\sin^{2}y} \times \sin2y\]
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lower limit will become 0
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I'm left with..
\[\LARGE \frac{1}{x}e^{\sin^{2}x+y} \times \sin2x\]
yrelhan4
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abbe yr.
x ki differentiation 1 hogi..
LH me numerator aur denominator alag alag diff krte hain.. :/
DLS
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maine bhi to wahi bola tha BC
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BUT 1 KA 0
yrelhan4
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to x ki diff 1 ho jayegi.. aur numerator me tune jo kra tha. dy/dx=0 krke aa jayega ans.
yrelhan4
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kya 1 ka 0? 1 baari diff krenge.. :/
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1/x hai to 1 ka karenge to 0 x ka karenge to 1
0/1=0
yrelhan4
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lol. 1*x ki diff. 0*x + 1*1.. lol
yrelhan4
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to upar sirf 1 thodi hai.. baaki bhi to hai. usse multiply krega saara.. to 1* ....... = ......... aa jayega.. :O
DLS
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badiya hai :|
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par sin2x kahan gaya?
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main question :P
yrelhan4
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newton lebinitz me integrand ki differentiation nhi krte.. limits ko integrand me daalte hain multiply by limits ki diff..
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medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal medal :*
yrelhan4
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:')
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seekho @shubhamsrg
yrelhan4
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lol. meri bezti badi achi krte ho aap log. :/
shubhamsrg
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maine upar kuch padha bhi nahi kya likha hai :3
knock
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What's the answer then? xD
shubhamsrg
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@yrelhan4 is the answer to all the questions . O:)
yrelhan4
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:')
knock
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Great xD I thought in a way to solve it but if it's already solved then ok (: