Question

is the integral of a function always talking about the area from 0 to some value.

Answer 1

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

Answer 2

how can you tell the difference?

Answer 3

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

Answer 4

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

Answer 5

that means that f(t) is the derivative of g(x). when it says g(4) that means that 4 becomes your x, or upper bound

Answer 6

can you take a look at that

Answer 7

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

Answer 8

g"(4)=f'(4)

Answer 9

so how come you cant just look at the picture and get the value of F(4) then subtract it with the value of f(0)

Answer 10

the problem i had was finding g(4)

Answer 11

I thought that you could take the f(4) then just subtract f(0) for g(4) but then...I got 2, but it was 3. So really it was the the area of f(1) to f(4) then subtract the area of f(0) to f(1)

Answer 12

the integral requires you tyake the antiderivative before you can plug in values. since f(4) is not a equation defines function for us, the best way is to use geometry to find the area of the graph of f(0) to f(4) BECAUSE g(4) is the integral of f(x) from 0 to 4, so youre just finding the area underneath the line of f(x) from 0 to 4

Answer 13

just count how many unit blocks fit under the line from f(0) to f(4), there should be no equations involved in this process unless its the area of a triangle and a square

Answer 14

ohh yeah....i forgot about that...you need to take the antiderivative before plugging in the values..that makes sense...so since you can't you have to just look at the areas....thanks you

Answer 15

bingo, good luck, click my fan button :)

Answer 16

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