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

embizze | High School Teacher | (Level 1) Educator Emeritus

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The first Fundamental Theorem of Calculus: If a functionf 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:

lfryerda | High School Teacher | (Level 2) Educator

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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)