- #1
ryt
- 9
- 0
i start to study integrals, i couldn't understand some things.
definite integrals i didnt understand, ill show it on some eg. on v as derivative of disaonce s.
the eg. was to calc how many distance s = ? does a particle travell when we throw it in air (up) and v reaches 0.
so in time t = 0 we have v0 = 10 m/s. The v(t) is a linear func decreasing and reaches 0 at time t1.
So the distance s is the definite integral of v(t) on intervals t0 and t1, so it is the surface under the v(t) from t0 to t1.
Im ok since here, i wondered where is this integral on a s(t) graph, i asked my friend and he told me theat this surface is at t1 as s(t1). Then i wondered how would i find some s(t) at some time between t0 and t1, and ocured to me i would take definite integral from t0 to some time t<t1, i would calc it and i would have some s(t<t1).
But now what i don't understand is what would happen if i move t0 to some t0<0, and then calc definite integral from t0<0 to t1 and what would i get??
some other thing i don't understand is:
in my book it says the derivative od definite integral is f(x)
[tex] \frac{d}{dx} \int_{a}^{x} f(t)dt = f(x) [/tex]
isnt it when we take definite integral of same func we get a number (rapresenting the surface under the function), so the derivative of a number is 0 not f(x)
definite integrals i didnt understand, ill show it on some eg. on v as derivative of disaonce s.
the eg. was to calc how many distance s = ? does a particle travell when we throw it in air (up) and v reaches 0.
so in time t = 0 we have v0 = 10 m/s. The v(t) is a linear func decreasing and reaches 0 at time t1.
So the distance s is the definite integral of v(t) on intervals t0 and t1, so it is the surface under the v(t) from t0 to t1.
Im ok since here, i wondered where is this integral on a s(t) graph, i asked my friend and he told me theat this surface is at t1 as s(t1). Then i wondered how would i find some s(t) at some time between t0 and t1, and ocured to me i would take definite integral from t0 to some time t<t1, i would calc it and i would have some s(t<t1).
But now what i don't understand is what would happen if i move t0 to some t0<0, and then calc definite integral from t0<0 to t1 and what would i get??
some other thing i don't understand is:
in my book it says the derivative od definite integral is f(x)
[tex] \frac{d}{dx} \int_{a}^{x} f(t)dt = f(x) [/tex]
isnt it when we take definite integral of same func we get a number (rapresenting the surface under the function), so the derivative of a number is 0 not f(x)