Without keeping Wolfram Alpha in your kitchen, solve it!
Ω = ∫ 0 1 ( 1 + x 2 0 1 5 ) 2 0 1 5 2 0 1 6 d x
Find the value of Ω 2 0 1 5 .
The answer is 0.500.
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2 solutions
Solved in almost same way.
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Where did you change paths? Because you said almost.
Oh, Fine then.Cheers.
just wrote 1/t as ln(t) with limits
Then solved the last value with calculator. Your way is better. Sometimes simple stuff dont come in my mind. LOL
Ah! Yours is much better! I did it using 2 substitutions x 2 0 1 5 = u , 1 + u 1 − u = v
same solution @Vishwak Srinivasan
Ω = ∫ 0 1 ( x 2 0 1 5 + 1 ) 2 0 1 5 2 0 1 6 1 d x Ω = ∫ ∞ 1 ( ( u 1 ) 2 0 1 5 + 1 ) 2 0 1 5 2 0 1 6 − u 2 d u = ∫ 1 ∞ ( 1 + u 2 0 1 5 ) 2 0 1 5 2 0 1 6 u 2 0 1 4 d u Ω = 2 0 1 5 1 ∫ 2 ∞ u 2 0 1 5 − 2 0 1 6 d u = − u 2 0 1 5 − 1 ∣ ∣ ∣ 2 ∞ Ω = ( 2 1 ) 2 0 1 5 1 Ω 2 0 1 5 = 2 1 substitute u = x 1 substitute u = 1 + u 2 0 1 5
Sorry for the image solution, too many fractions.