# How would you use the fundamental theorem to integrate the definite integral e^(1/x)/x^2 dx on interval 1 to 1/2?

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The first Fundamental Theorem of Calculus: If a function`f` is continuous on theclosed interval `[a,b]` and `F` is the antiderivative of `f` on the interval `[a,b]` , then `int_a^bf(x)dx=F(b)-F(a)` .

Here `f(x)=(e^(1/x))/x^2` is continuous on [1/2,1] so the Fundamental Theorem applies. The antiderivative of `(e^(1/x))/x^2` is `-e^(1/x)` so:

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`int_1^(1/2)(e^(1/x))/x^2dx=-e^(1/((1/2)))-(-e^1)=-e^2+e=e(1-e)`

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**Sources:**

To integrate `\int_1^{1/2}{e^{1/x}}/{x^2}dx`, use the substitution `u=1/x`.

Then `du=-1/x^2 dx` and when `x=1`, `u=1` and when `x=1/2`, `u=2`.

This means that the integral becomes

`-\int_1^2e^udu`

`=-e^u|_1^2`

`=-(e^2-e^1)`

`=-e(e-1)`

`=e(1-e)`