Given: $\displaystyle \int _1^{x}f(t)dt = x^2 - 2x +1$ Find $\displaystyle f(x)$
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Originally Posted by penguinpwn Given: $\displaystyle \int _1^{x}f(t)dt = x^2 - 2x +1$ Find $\displaystyle f(x)$ ... take the derivative of both sides of the equation w/r to x
So would that lead to $\displaystyle \int _1^{x}f(t)dt = x^2 - 2x +1$ $\displaystyle \frac {dt} {dx} = (2x - 2)dx $ Then what?
Originally Posted by penguinpwn So would that lead to $\displaystyle \int _1^{x}f(t)dt = x^2 - 2x +1$ $\displaystyle \frac {dt} {dx} = (2x - 2)dx $ Then what? are you not familiar with the Fundamental Theorem of Calculus? ... $\displaystyle \frac{d}{dx} \int_a^x f(t) \, dt = f(x)$
I'm relatively new to the subject, but yes. I just don't understand how it's applied here
$\displaystyle \frac{d}{dx} \left(\int _1^{x}f(t) \, dt = x^2 - 2x +1\right)$ $\displaystyle f(x) = 2x-2$ that's all there is to it.
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