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Integration of a curve in the 2nd quadrant.
Hi, The graphs of y = 4 and y = 1 - x meet at (-3, 4). The required area will be (area from y = 4 down to the x-axis) MINUS (area from y = 1 - x down to x-axis), starting at x = -3 and ending at x = 0. So, your calculation will be INTEGRAL (-3 to 0) of 4 MINUS INTEGRAL (-3 to 0) of 1...- RickB
- Post #3
- Forum: Mathematics Advanced
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