Happy Birthday, Mehul!
1 ∫ 1 5 e x sinh ( x ) d x
If the above integral has a closed form: d a − e b + e c for some integers a , b , c , d where a < 0 , then find the value of a + b + c + d .
Clarification: e denotes the Euler's number , e ≈ 2 . 7 1 8 2 8 .
This problem is original and is dedicated to Mehul on his birthday.
The answer is 8.
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2 solutions
We posted same solutions at same time :P
I was too lazy to post a solution , hence relied on you to post a solution :P
BTW I did by same method.
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:-)..... Thanks for relying on me and thinking me 'that' capable ;-)
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Why don't you post good calculus problems? We juniors would definitely benefit from it :)
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@Nihar Mahajan – Yes please post bro! @Rishabh Cool
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@Harsh Shrivastava – It might be the starting... Please make me know if you spot any error ;-)
@Nihar Mahajan – Easy one to start with .... Surely in future I'll try to post some really difficult ones after learning some functions.. ;-)
Same way !!! It was a hard one for me though :>
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What is your real name?
The Cool-Man strikes again! Same solution here.
Using sinh ( x ) = 2 e x − e − x ,
our integrand transforms into ∫ 1 1 5 [ 2 e 2 x − 2 1 ] d x
which can evaluated using standard techniques and comes out to be 4 − 2 8 − e 2 + e 3 0
Using sin h x = 2 e x e 2 x − 1 (See this ) I = ∫ 1 1 5 e x sinh ( x ) d x = ∫ 1 1 5 2 e 2 x − 1 = 4 1 ( e 2 x − 2 x ) ∣ 1 1 5 = 4 − 2 8 − e 2 + e 3 0