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If $ f $ is continuous on $ \mathbb{R} $, prove that$$ \int^b_a f(-x) \, dx = \int^{-a}_{-b} f(x) \, dx $$For the case where $ f(x) \ge 0 $ and $ 0 < a < b $, draw a diagram to interpret this equation geometrically as an equality of areas.

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01:56

Frank Lin

Calculus 1 / AB

Chapter 5

Integrals

Section 5

The Substitution Rule

Integration

Harvey Mudd College

University of Nottingham

Idaho State University

Boston College

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.

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okay, We know that we're gonna have our limits of integration from negative beat too Negative A Because, remember, from A to B f of x, T X is equivalent to negative from B to a FX DX. It's the same thing. So we know now that we can draw our diagram for this question, which is what they asked us to. D'oh! We know this is f of AKs and this is off of negative X recall the fact that half of negative X is a reflection off Michael's FX and the Y axis.

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