Recall the interpretation of the definite integral as the signed area under the curve.
Portions below the graph count as "negative area" even though such a thing could not
exist geometrically.
f (x)dx
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= 3 - 1 + 2 = 4 |
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f (x)dx
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= 1 |
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f (x)dx
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= 3 - 1 + 2 - 5 = - 1 |
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