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In Exercises $30-33$ find the median.pd $f$ in Exercise 28.

0.28311

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 8

Applications of the Definite Integral

Integrals

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Lectures

05:53

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.

40:35

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.

01:57

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07:02

the graphical understanding of of an average value makes perfect sense. If you were to examine what I mean. So let me just show you the graph of y equals X plus two. So it starts at 02 and then over here this point is going to be Than one arrow here. 46 because if you're plugging for for ex you get up to six um Yeah, so if we imagine what this is, the average value saying, hey if we find the area under this curve, what I want to do is make a identical rectangle that has the same area. So basically I want to take this area like cut it off and fold it over here and what's going to be the height of that new rectangle? Well it's going to be for um now if you do the arithmetic or you do one over the upper bound minus the lower bound, the integral from 0 to 4 of the quality of X-plus two dx you can find that the area under the curve. It's a trapezoid. Uh Probably the fast way of doing this, otherwise you can actually evaluate being enroll or you add one to the exponents, multiply by the reciprocal of the exponents. And what I'm doing is I'm coming up with the derivative, the derivative of this gets back to here and if you plug in your balance four squared is 16, half of 16 is eight, two times four. Mhm. It was a silly mistake, I just changed the sign of this. Uh Anyway. Uh You do get the correct answer of 16 divided by four, which is four.

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