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Alem2000
Sep25-04, 03:38 PM
Hello page...im taking integral calculus and we are past integration of improper integrals. I know how to use the fundamental theorm but i dont get the first part... \frac{d}{dx}\int^x_af(t)dt=f(x)

the book used it in an example....find the dirivative of g(x)=\int_0^1\sqrt{1+t^2}dt....the book goes on to tell you the answer but it show NO STEPS...it is James Stewart Calculus 2nd edition i believe if anyone has the same book..page 383..but anyway can some one go through the steps...plz :redface:

mathman
Sep25-04, 04:02 PM
There is something wrong in the statement of the problem. You have g(x) = something, where x does not appear. As stated g'(x)=0.

Alem2000
Sep25-04, 05:15 PM
sorry I copied down the problem wrong...this is the correct one


g(x)=\int_0^x\sqrt{1+t^2}dt

mathwonk
Sep25-04, 06:09 PM
just plug into the statement you gave of the ftc. this is a special case. i.e. what is f(t) here?

Tide
Sep25-04, 06:13 PM
We can get it from first principles:
\frac {dg}{dx} = \lim_{h \rightarrow 0} \frac {\int_0^{x+h} \sqrt{1+t^2} dt - \int_0^{x} \sqrt {1+t^2} dt}{h}
\frac {dg}{dx} = \lim_{h \rightarrow 0} \frac {\int_{x}^{x+h} \sqrt{1+t^2} dt}{h}
\frac {dg}{dx} = \lim_{h \rightarrow 0} \sqrt{1+x^2} \frac {\int_{x}^{x+h} dx}{h} = \sqrt {1+x^2}

mathwonk
Sep25-04, 07:55 PM
what "first principle" did you use in the next to last step?

JasonRox
Sep26-04, 11:54 AM
I got the fifth edition, so I can't help you there. I just hope I don't encounted similiar problems.

matt grime
Sep26-04, 12:18 PM
what problems do you think the OP encountered in the book? the solution is self evident and any problems the OP had are nothing to do with the book. the book has many faults, if it's the one i think it is, but that isn't one of them.