What is ∫ 1 3 e 1 x x 2 d x ?How do you approach solving this?

Question

What is      ∫    1    3              e                        1          x                            x              2              d  x  ?How do you approach solving this?

faux0101d · Accepted Answer

In general:  ∫      f    ′    (  x  )        e          f      (      x      )          d  x  =      e          f      (      x      )        +  Cso in our case  ∫            e              1                  /                x                    x      2          d  x  =  −  ∫      (    −          1              x        2                  d    x    )        e          1              /            x        =  −      e          1              /            x        +  Cand with limits:                          ∫        1        3                              e                      1                          /                        x                                    x          2                          d      x      =      −              e                  1                      /                    x                    |        1    3    =  −  (      e          1              /            3        −  e  )  =  e  −      e    3

mafalexpicsak · Accepted Answer

Using Taylor Series is a good way to approximate definite integrals if the integrand is difficult to work with (though in your case, it is not.) An example would be

\int_{1}^{3} \frac{e^{- x^{2}}}{x} d x .

Recall that

\begin{matrix} (1) & e^{x} = \sum_{k = 0}^{\infty} \frac{x^{k}}{k!} Taylor Series for e^{x} \end{matrix}

Thus,

\begin{matrix} (2) & e^{1 / x} = \sum_{k = 0}^{\infty} \frac{{(\frac{1}{x})}^{k}}{k!} \end{matrix}

\begin{matrix} (2) & e^{1 / x} = \sum_{k = 0}^{\infty} \frac{{(\frac{1}{x})}^{k}}{k!} \end{matrix}

Writing out the series expansion (the first few terms) for

e^{1 / x}

yields

\begin{matrix} (3) & 1 + \frac{1}{x} + \frac{1}{2 x^{2}} + \frac{1}{6 x^{3}} + \frac{1}{24 x^{4}} + O {(\frac{1}{x})}^{5} \end{matrix}

where

O {(\frac{1}{x})}^{5}

is the higher order terms. Now our original integral can be rewritten as

\begin{matrix} (4) & \int_{1}^{3} \frac{1 + \frac{1}{x} + \frac{1}{2 x^{2}} + \frac{1}{6 x^{3}} + \frac{1}{24 x^{4}}}{x^{2}} d x \end{matrix}

This simplifies to

\begin{matrix} (5) & \int_{1}^{3} \frac{1}{x^{2}} + \frac{1}{x^{3}} + \frac{1}{2 x^{4}} + \frac{1}{6 x^{5}} + \frac{1}{24 x^{6}} d x \end{matrix}

Integrate each piece and you get approximately 1.32106. Notice that the error between this answer and the other answer is small because

e - e^{1 / 3} \approx 1.322669

. If we used more terms in the Taylor Series expansion, the error would have been even smaller.

What is int_1^3 e^(1/x)/x^2 dx ? How do you approach solving this ?

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