Evaluate the following, int_0^1 (x^e +e^x) dx. I know the anti-derivative of x^e is x^(e+1)/(e+1). How would i solve this,would i use F(b) +F(a)?

Jan 15, 2017

I got $\frac{e}{e + 1} + e - 1$, however the answer is $\frac{1}{e + 1} + e - 1$,

${\int}_{0}^{1} \left({x}^{e} + {e}^{x}\right) \setminus \mathrm{dx} = \frac{1}{e + 1} + e - 1$

Explanation:

I did F(1)-F(0) and i did anti-derivative of x^e * derivative of x^e which is $e {x}^{e - 1}$

${\int}_{0}^{1} \left({x}^{e} + {e}^{x}\right) \setminus \mathrm{dx} = {\left[{x}^{e + 1} / \left(e + 1\right) + {e}^{x}\right]}_{0}^{1}$
$\text{ } = \left\{\left({1}^{e + 1} / \left(e + 1\right) + {e}^{1}\right) - \left({0}^{e + 1} / \left(e + 1\right) + {e}^{0}\right)\right\}$
$\text{ } = \left\{\left(\frac{1}{e + 1} + e\right) - \left(0 + 1\right)\right\}$
$\text{ } = \frac{1}{e + 1} + e - 1$

Where we used:

${1}^{n} = 1 \forall n \in \mathbb{R} \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \implies {1}^{e + 1} = 1$
${0}^{n} = 0 \forall n \in \mathbb{R} - \left\{0\right\} \setminus \implies {0}^{e + 1}$
${x}^{n} = 1 \forall n \in \mathbb{R} \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \implies {e}^{0} = 1$