Question

Find the derivative f’(x)

Answer 1

@Meepi can u help?

Answer 2

There's a shortcut to doing these... the derivative essentially undoes the integral, you just have to take the derivative of the limit when it's some function of x (in this case x so it'll just be 1).

\[\Large \frac{ d }{ dx } \int\limits_{0}^{x} (t^2 - 3t +2) dt = x^2 - 3x+2\]

Answer 3

Or you can do it the long way \[\Large \frac{ d }{ dx } \int\limits\limits\limits_{0}^{x} (t^2 - 3t +2) dt = \frac{ d }{ dx}\left[ \frac{ t^3 }{ 3 } -\frac{ 3t^2 }{ 2 } +2t \right]_{0}^{x} = \]Now plug in the limits, which gives (since plugging in 0 just gives 0)
\[\Large \frac{ d }{ dx }\left[ \frac{ x^3 }{ 3 } -\frac{ 3x^2 }{ 2 } +2x \right] = x^2 - 3x +2\]

Answer 4

okay so the final answer is x^2 - 3x + 2?

Answer 5

Yep. Note that if the limit is something like x^2 or 2x, then you have to actually differentiate that part and bring it out out the front, after plugging in the limit, eg if it's x^2, derivative is 2x so:
\[\Large \frac{ d }{ dx } \int\limits\limits\limits_{0}^{x^2} (t^2 - 3t +2) dt = (2x) \left[ (x^2)^2 - 3(x^2) +2\right]\]

Answer 6

ohhh ok got you thanks!

Answer 7

And the long way would give the same thing\[\Large \frac{ d }{ dx}\left[ \frac{ t^3 }{ 3 } -\frac{ 3t^2 }{ 2 } +2t \right]_{0}^{x^2} = \frac{ d }{ dx}\left[ \frac{ x^6 }{ 3 } -\frac{ 3x^4 }{ 2 } +2x^2 \right] = \] \[\Large 2x^5 - 6x^3 + 4x\] which is equal to (if you expand this out)\[\Large (2x) \left[ (x^2)^2 - 3(x^2) +2 \right]\]

Answer 8

You can do this with any kind of function... eg for this, the limit is 2x, so derivative is 2:\[\Large \frac{ d }{ dx} \int\limits_{0}^{2x}\sin t dt = 2 \sin (2x) \] cancel out the integral/derivative, plug in the limit 2x and multiply by the derivative of 2x which is 2.

Answer 9

Okay

Answer 10

Okay