## anonymous 5 years ago is the integral of a function always talking about the area from 0 to some value.

1. anonymous

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

2. anonymous

how can you tell the difference?

3. anonymous

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

4. anonymous

$\int\limits_{lowerbound}^{upperbound}$

5. anonymous

6. anonymous

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

7. anonymous

can you take a look at that

8. anonymous

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

9. anonymous

g"(4)=f'(4)

10. anonymous

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)

11. anonymous

the problem i had was finding g(4)

12. anonymous

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)

13. anonymous

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

14. anonymous

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

15. anonymous

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

16. anonymous

bingo, good luck, click my fan button :)

17. anonymous

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