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WaynexBest ResponseYou've already chosen the best response.0
The topic is covered in single variable calculus. The MIT course I recommend you take a look at is 18.01.
 one year ago

anonymousssBest ResponseYou've already chosen the best response.0
The area under a curve. Such as the area under the curve of y = x from 1 to 0 = 1 * 1 * 0.5 = 0.5 while another solution is \[\int\limits_{0}^{1}(x) dx = 1^2 \div 2  0^2 \div 2 = 1 \div 2  0 \div 2 = 0.5  0 = 0.5\]
 one year ago
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