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Let $ f(x) = 0 $ if $ x $ is any rational number and $ f(x) = 1 $ if $ x $ is any irrational number. Show that $ f $ is not integrable on $ [0, 1] $.
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$f$ is not integrable on [0,1].
Calculus 1 / AB
The Definite Integral
Oregon State University
In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.
In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.
Let $f(x)=0$ if $x$ is any…
Prove that f is defined as…
We know that the interval I over and comma I plus one over and will always have a rational and irrational number regardless of how large end becomes as it continues on. Therefore, we know we have the limit As one approaches infinity. We know this is one and then we know the same thing. But when we have zero times one over end a zero. As you can see, the limit depends on the choice. Therefore, off is not possible to be integrated. It's not Intercable.
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