∫102xex2−1dx\int _0^1\:2xe^{x^2-1}dx [on hold]

for the answer I’m getting:

1+1e1+\frac{1}{e}

but according to symbolab, I get:

1−1e1-\frac{1}{e}

steps I’m currently taking with u substitution; I get:

∫eu\int \:e^u

then plugging back u;

ex2−1e^{x^2-1} (from 0 to 1)

then I compute,

e−1+e0e^{-1}+e^0 = 1+1e1+\frac{1}{e}

I’m not sure

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1 Answer
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For the definite integral it would be e1−1−e0−1e^{1-1} – e^{0-1}

  

 

ok ha I just saw that i see
– Temur Bakhriddinov
Oct 21 at 1:17