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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.

## $$\int_{-b}^{-a} f(x) d x$$

Integrals

Integration

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### Video Transcript

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.

Integrals

Integration

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