The formula for the length from a to b of a continuos equation, whose derivative is also continous is:
length = [tex] \int\limits^a_b { \sqrt{1+ (f '(x))^{2} } } \, dx [/tex]
Here, f '(x) = e^(-x)
a = 0 and b = 2
An the required integral is:
integral of {1+ [e^(-x )]^2}dx from 0 to 2
which is x - e^(-2x) / 2 from 0 to 2 =
2 - e^(-4) / 2 - 0 + e^(0) / 2 = 2 - e^(-4)) / 2 + 1/2 = 5/2 - e^(-4) /2 ≈ 2.49
Answer: 2.49
