Hello Everyone! Today We are asked to find $\displaystyle \int_{1}^{2} e^\frac{1}{x}dx$. We are given two option. First is by evaluating the definite integral itself or we can do numerical approximation using both Trapezoidal and Mid point Rules with sub interval n valued at 10. Since I've figured out that integrating such integral would be a , I just gone on and used the Numerical Approximation way. Everything is well until I am asked to find the error for both $\displaystyle T_{10}$ and $\displaystyle M_{10}$. I know the formula and the value of for the second derivative of the function $\displaystyle e^\frac{1}{x}$ to use it as a rule to find constant $\displaystyle K$ from the formula $\displaystyle |E_T|\leq\frac{K(b-a)^3}{12n^2}$ and $\displaystyle |E_M|\leq\frac{K(b-a)^3}{24n^2}$. Now I am stuck at how to find K when $\displaystyle K\geq|\frac{e^\frac{1}{x}(2x+1)}{x^4}|$ at interval $\displaystyle [1,2]$. Anyone got any idea to find $\displaystyle K$? Thanks y'all in advance!

Edit: I have an idea on how to solve this one. As we can see, x is bounded in [1,2], and that 1/x would only max out at x=1 since anything more than 1 would make x less so we can conclude that $\displaystyle \frac{1}{x}\leq 1$. Since K is the max value of the second derivative, then max value is acquired when 1/x is evaluated at x=1. So that $\displaystyle K\geq|(e^\frac{1}{x}(\frac{2}{\frac{1}{x}}+1)) (\frac{1}{x})^4|$ so if we substitute 1 to all 1/x, we get: $\displaystyle K=3e$. Is that the right way to do it?Hehe it is late in the morning and I am still pumped to do this problem!