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mathsloverBest ResponseYou've already chosen the best response.0
It is easy , it just looks long. I will do that after lunch, ok?
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
\[\Huge \lim_{x \rightarrow 0}( \frac{1}{x} \int\limits_{0}^{a} e^{\sin^{2}t}dt\int\limits_{x+y}^{a} e^{\sin^{2}t}dt)\]
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
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.
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
sorry the term 1/x is outside everything
 11 months ago

yrelhan4Best ResponseYou've already chosen the best response.2
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..
 11 months ago

yrelhan4Best ResponseYou've already chosen the best response.2
now combining the two limits.. 0 to a and a to x+y .. 0 to x+y ..
 11 months ago

yrelhan4Best ResponseYou've already chosen the best response.2
because the integrand is same..
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
0 to a and 0 to x+y=0 to x+y
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
\[\Huge \frac{1}{x} e^{\sin^{2}x+y}\]
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
differenatiate and bingo :O
 11 months ago

yrelhan4Best ResponseYou've already chosen the best response.2
why did you not do anything to 1/x ?
 11 months ago

yrelhan4Best ResponseYou've already chosen the best response.2
multiplication theorem.. hmm..
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
1/x was outside the integral sorry :P
 11 months ago

yrelhan4Best ResponseYou've already chosen the best response.2
aur jab 0 put krenge.. to bhi to kuch aayega. vo kahan gyi? :P
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
\[\Huge \lim_{x \rightarrow 0} \frac{1}{x}(\int\limits_{y}^{a}e^{\sin^{2}t} dt\int\limits_{x+y}^{a}e^{\sin^{2}t} dt)\]
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
\[\Large \lim_{x \rightarrow 0} \frac{1}{x}(\int\limits_{y}^{x+y}e^{\sin^{2}t} dt)\] On changing the sign and revevrsing the limits and combining them
 11 months ago

yrelhan4Best ResponseYou've already chosen the best response.2
acha vo 'a' vala to 0 ho jayega.. a ki differentiation 0.. but hamne x+y ko differentiate bhi nhi kra.. :/
 11 months ago

shubhamsrgBest ResponseYou've already chosen the best response.0
kya hua? kya dikkat hai ? (:
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
\[\Huge e^{\sin^{2}{x+y}}(1+\frac{dy}{dx})e^{\sin^{2}y} \frac{dy}{dx}\]
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
\[\Huge e^{\sin^{2}x+y}\] answer
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
y is behaving as a constant,we assume y independent of x
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
idk what i wrote please someone explain :P
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
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
 11 months ago

yrelhan4Best ResponseYou've already chosen the best response.2
limit y se x+y ?? how??
 11 months ago

yrelhan4Best ResponseYou've already chosen the best response.2
i said 0 to x+y.. :/
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
but still didnt get it :
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
\[\LARGE e^{\sin^{2}x+y}\] differentiating this^
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
\[\LARGE e^{\sin^{2}x+y} \times \sin2x(+0)\]
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
\[\LARGE e^{\sin^{2}y}\] Differentiating the lower limit^
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
\[\LARGE e^{\sin^{2}y}=>e^{\sin^{2}y} \times \sin2y\]
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
lower limit will become 0
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
I'm left with.. \[\LARGE \frac{1}{x}e^{\sin^{2}x+y} \times \sin2x\]
 11 months ago

yrelhan4Best ResponseYou've already chosen the best response.2
abbe yr. x ki differentiation 1 hogi.. LH me numerator aur denominator alag alag diff krte hain.. :/
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
maine bhi to wahi bola tha BC
 11 months ago

yrelhan4Best ResponseYou've already chosen the best response.2
to x ki diff 1 ho jayegi.. aur numerator me tune jo kra tha. dy/dx=0 krke aa jayega ans.
 11 months ago

yrelhan4Best ResponseYou've already chosen the best response.2
kya 1 ka 0? 1 baari diff krenge.. :/
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
1/x hai to 1 ka karenge to 0 x ka karenge to 1 0/1=0
 11 months ago

yrelhan4Best ResponseYou've already chosen the best response.2
lol. 1*x ki diff. 0*x + 1*1.. lol
 11 months ago

yrelhan4Best ResponseYou've already chosen the best response.2
to upar sirf 1 thodi hai.. baaki bhi to hai. usse multiply krega saara.. to 1* ....... = ......... aa jayega.. :O
 11 months ago

yrelhan4Best ResponseYou've already chosen the best response.2
newton lebinitz me integrand ki differentiation nhi krte.. limits ko integrand me daalte hain multiply by limits ki diff..
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
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 :*
 11 months ago

yrelhan4Best ResponseYou've already chosen the best response.2
lol. meri bezti badi achi krte ho aap log. :/
 11 months ago

shubhamsrgBest ResponseYou've already chosen the best response.0
maine upar kuch padha bhi nahi kya likha hai :3
 11 months ago

knockBest ResponseYou've already chosen the best response.0
What's the answer then? xD
 11 months ago

shubhamsrgBest ResponseYou've already chosen the best response.0
@yrelhan4 is the answer to all the questions . O:)
 11 months ago

knockBest ResponseYou've already chosen the best response.0
Great xD I thought in a way to solve it but if it's already solved then ok (:
 11 months ago
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