Evaluate the integral of (e^y+y^e)dy from 0 to 1
∫01(ey+ye)dy=ey+ye+1e+1∣01=e+1e+1−1==(e−1)(e+1)+1e+1=e2−1+1e+1=e2e+1\int_0^1(e^y+y^e)dy = e^y+\cfrac{y^{e+1}}{e+1}|_0^1 = e + \cfrac{1}{e+1} - 1= \\ = \cfrac{(e-1)(e+1)+1}{e+1} = \cfrac{e^2-1+1}{e+1} = \cfrac{e^2}{e+1}∫01(ey+ye)dy=ey+e+1ye+1∣01=e+e+11−1==e+1(e−1)(e+1)+1=e+1e2−1+1=e+1e2
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