For the function Ftlnt let ftFt Write the integral ab ftdt
For the function F(t)=lnt , let f(t)=F\'(t). Write the integral a-->b f(t)dt and evaluate it with the Fundamental Theorem of Calculus.
Integral 1-->2 _______? dt=_______?
Please take not of how I have established the question here.
(Note that your answer must be exact, not a decimal approximation.)
Integral 1-->2 _______? dt=_______?
Please take not of how I have established the question here.
(Note that your answer must be exact, not a decimal approximation.)
Solution
Integral 1-->2 f(t) dt= ln(2)
this is because F\' dt= F
that is the integral and the derivative cancel out to the original function
So 12(ln(t))\' dt= ln(t)|12=ln(2)-ln(1)=ln(2)-0=ln(2)
since (ln(t))\'=1/t you might also have
Integral 1-->2 1/t dt= ln(2)
it depends what the tester wants
