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MIT 18.01 Single Variable Calculus (OCW) 9 Online
OpenStudy (anonymous):

Hi! I just haven't get why the answer to the minus integral of xdx is equal to -x^2/2? I mean, why do we have to divide by 2? Didn't quite get that on the previous lecture. Thanks.

OpenStudy (phi):

Let's say \[ y = - \frac{1}{2} x^2\] now take the derivative of both sides with respect to x: \[ \frac{dy}{dx}= -\frac{1}{2} 2 x \\ dy = - -\frac{1}{2} 2 x\ dx \\ dy= -x \ dx \] the idea is the derivative and the integral are inverse operations (up to a constant) thus the rule for the integral of x to some power: \[ \int x^n \ dx= \frac{1}{n+1} x^{n+1} + C \] and taking the derivative: \[ \frac{d}{dx} \int x^n \ dx= \frac{d}{dx} \left(\frac{1}{n+1} x^{n+1} + C\right)\\ = \frac{1}{n+1} \frac{d}{dx} x^{n+1} + \cancel{\frac{d}{dx}C}0\\ = \frac{1}{n+1} (n+1) x^{n+1-1} \\ =x^n \]

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