Question

`(e ^(x) + e^( - x) )`
how would one find the definite integral from the 0 to 1​

Answer 1

`\displaystyle \int^(1)_0 \left(e^x + e^(-x) \right) dx\\\\=\displaystyle \int^(1)_0 e^x ~~dx+\displaystyle \int^(1)_0 e^(-x) ~ dx\\\\=\left[ e^x\right]^(1)_(0) -\left[e^(-x) \right]^(1)_0~~~~~~~~~~~~~~~~~~~~~~~;\left[\displaystyle \int e^(mx) ~ dx = \frac 1m e^(mx) + C \right]\\\\=\left(e^1 - e^0\right) - \left(e^(-1) - e^0\right)\\\\=(e-1)-\left(\frac 1 e - 1\right)\\\\=e-1 -\frac 1 e+1\\\\=\frac{e^2-1}e`