The value of integral ₀^ 1 1 + e^x ,dx is...

Step-by-step solution

Let I = ₀^ 1 1 + e^x dx Multiplying the numerator and the denominator by e^ -x : I = ₀^ e^ -x e^ -x + 1 dx Let e^ -x + 1 = t Differentiating both sides, -e^ -x dx = dt e^ -x dx = -dt Limits of integration: When x = 0 , t = e^0 + 1 = 2 When x , t = e^ - + 1 = 1 Substituting these into the integral: I = ₂^1 -dt t = ₁^2 dt t I = [ t]₁^2 = 2 - 1 = 2 Answer: 2