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Justify (3) for the case $ h < 0 $.
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Calculus 1 / AB
The Fundamental Theorem of Calculus
Harvey Mudd College
University of Nottingham
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.
Solve each of the followin…
Solve each equation.$$…
Solve each equation. Write…
Solve each equation. Check…
Determine the value of $h(…
okay. We know the first thing we can look at is the extreme value theory, which says that there are two numbers we can just call him. You envy in this context in the range that that specifies that off of you is equivalent to, um and after v is capital at the's are absolute minimum and absolute maximum values. Therefore, what we have is g of excuse age minus G of axe divide by H, which is one over acts from acts to expose age off of t did to you. We know this is equivalent to so the problem asked us to justify So we're saying this is indeed equation 31 h is less than zero.
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