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Evaluate int (e^x+x^e) dx from 1 to 0I'll mark your answer brainliest, please answer fast​

Answer»

ANSWER:

\\\tt 1 - e - \dfrac{1}{e+1}

EXPLANATION:

\\\tt \int _1 ^0 (e^x + x^e) dx \\ \\\tt = \int _1 ^0 e^x dx + \int _1 ^0 x^edx \\ \\\tt = [e^x]_1^0 + [\dfrac{x^{e+1}}{e+1} ]_1^0 \\ \\\tt = (1-e ) + (\dfrac{0^{(e+1)}}{e+1} - \dfrac{1^{(e+1)}}{e+1} )\\ \\\tt = 1 - e - \dfrac{1}{e+1}

FORMULAS USED:

\\\tt \int_a^b e^x dx = [e^x]_a^b = e^b -e^a\\ \\\\\tt \int_a^b x^y dx = [\dfrac{x^{y+1}}{y+1} ]_a^b = (\dfrac{b^{y+1}}{y+1} )- (\dfrac{a^{y+1}}{y+1} )


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