Question

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

Answer 1

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:

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

Answer 2

⠀⌬⠀ Evaluate :

$\qquad \qquad \bigstar \:\:\sf \int_1^0 \:\big( \:e^x \: + \:x^e\: \big)\:dx\:$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

$\qquad \dashrightarrow \sf \int_1^0 \:\big( \:e^x \: + \:x^e\: \big)\:dx\:\:\:$

$\qquad \qquad \bigstar \:\:\underline{\purple {\pmb{\sf \:\: By \:\:Evaluating\:the\:Given \:\: ,\:we\:get\:;}}}$

$\qquad \dashrightarrow \sf \int_1^0 \:\big( \:e^x \: + \:x^e\: \big)\:dx\:\:\:\\\\\\\qquad \dashrightarrow \sf \int_1^0 \:\big( \:e^x \: + \:x^e\: \big)\:dx\:\:\:\\\\ \\ \qquad \dashrightarrow \sf \int_1^0 \: \:e^x \:dx\: + \: \int_1^0\:x^e\:\:dx\:\:\:\\\\\\\qquad \dashrightarrow \sf \: \:\big\{\:e^x\big\}^0_1 \:\: + \: \Bigg[ \dfrac{x^{e+ 1 }}{e + 1 }\:\Bigg]_1^0 \:\:\:\:\\\\\\\qquad \dashrightarrow \sf \: \:\big\{\:e^0 \:-\:e^1\big\} \:\: + \: \Bigg[ \dfrac{x^{e+ 1 }}{e + 1 }\:\Bigg]_1^0 \:\:\:\:\\\\ \\ \qquad \dashrightarrow \sf \: \:\big\{\:1 \:-\:e\big\} \:\: + \: \Bigg[ \dfrac{x^{e+ 1 }}{e + 1 }\:\Bigg]_1^0 \:\:\:\:\\\\\\\qquad \dashrightarrow \sf \: \:\big\{\:1 \:-\:e\big\} \:\: + \: \Bigg[ \dfrac{0^{e+ 1 }}{e + 1 }\:-\:\dfrac{1^{e+ 1 }}{e + 1 }\:\Bigg] \:\:\:\:\\\\ \\ \qquad \dashrightarrow \sf \: \:\big\{\:1 \:-\:e\big\} \:\: + \: \Bigg[ \dfrac{0^{e+ 1 }}{e + 1 }\:-\:\dfrac{e+ 1 }{e + 1 }\:\Bigg] \:\:\:\:\\\\ \\ \qquad \dashrightarrow \sf \: \:\big\{\:1 \:-\:e\big\} \:\: + \: \Bigg[ \dfrac{0^{e+ 1 }}{e + 1 }\:-\:\cancel {\:\dfrac{e+ 1 }{e + 1 }}\:\Bigg] \:\:\:\:\\\\ \\ \qquad \dashrightarrow \sf \: \:\big\{\:1 \:-\:e\big\} \:\: + \: \Bigg[ \:\dfrac{0^{e+ 1 }}{e + 1 }\:\Bigg] \:\:\:\:\\\\ \\ \qquad \dashrightarrow \sf \: \:\big\{\:1 \:-\:e\big\} \:\: + \: \Bigg[ \:\dfrac{1}{e + 1 }\:\Bigg] \:\:\:\:\\\\ \\ \qquad \dashrightarrow \sf \: \:\:1 \:-\:e \:\: + \:\dfrac{1}{e + 1 } \:\:\:\:\\\\ \\ \qquad \dashrightarrow\underline {\boxed {\pmb{ \frak{\purple { \: \:\:1 \:-\:e \:\: + \:\dfrac{1}{e + 1 } \:}}}}}\:\:\:$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\qquad \qquad \underline {\bigstar\:\:\pmb{\mathbb{ ADDITIONAL \:\:INFORMATION \::\:}}}\:$

$\dag \:\:\underline { \underline {\purple{\sf Integrals \:\:-\: }}}$

$\qquad \qquad \boxed{\boxed{\begin{array}{cc}\displaystyle\circ\sf\:\int{1\:dx}=x+c\\\\\circ\sf\:\int{a\:dx}=ax+c\\\\\circ\sf\:\int{x^n\:dx}=\dfrac{x^{n+1}}{n+1}+c\\\\\circ\sf\:\int{sin\:x\:dx}=-cos\:x+c\\\\\circ\sf\:\int{cos\:x\:dx}=sin\:x+c\\\\\circ\sf\:\int{sec^2x\:dx}=tan\:x+c\\\\\circ\sf\:\int{e^x\:dx}=e^x+c\end{array}}}$