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nope, integrals can be improper, indefinite, and definite from values other than 0.

how can you tell the difference?

by the values above and below the integral symbol, known as the bounds

\[\int\limits_{lowerbound}^{upperbound}\]

can you take a look at that

g'(4) is the derivative at x=4, since we know that f(t) is our derivative then g'(4) =f(4)

g"(4)=f'(4)

the problem i had was finding g(4)

bingo, good luck, click my fan button :)

wait last question...so why do you need to subtract the area under from 0 to 1?