Definite integral help

ryt

10
0
i start to study integrals, i couldnt 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 dont 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 dont 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)
 

quasar987

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ryt said:
some other thing i dont 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)
Notice that the superior bound of the integral is x. So if f(x) has a primitive (i.e. if there exists a function F(x) such that F'(x)=f(x)), then [itex]\int_a^x f(t) dt[/itex]=F(x)-F(a). You can now differentiate this with respect to x, and you get just f(x), since F'(x)=f and F'(a)=0.

(Note that as soon as f is continuous on (a,b), it has a primitive on (a,b))
 
Last edited:

ryt

10
0
thx

how do u know F'(a)=0 ??
 
536
2
He meant differentiate f(a) (which is a number) with respect to x, not evaluate f'(x) at a.
 

HallsofIvy

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quasar987 said:
Notice that the superior bound of the integral is x. So if f(x) has a primitive (i.e. if there exists a function F(x) such that F'(x)=f(x)), then [itex]\int_a^x f(t) dt[/itex]=F(x)-F(a). You can now differentiate this with respect to x, and you get just f(x), since F'(x)=f and F'(a)=0.

(Note that as soon as f is continuous on (a,b), it has a primitive on (a,b))
No, F'(a) is not 0, it is f(a). What you may have meant to say is that [itex]F(a= \int_a^a f(t)dt)= 0[/itex].
 

quasar987

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?!

But Halls, if you agree that that if F is a primitive of f, then the derivative of the function G(x) =[itex]\int_a^x f(t) dt=F(x)-F(a)[/itex], is F'(x) + [itex]\frac{d}{dx}F(a)[/itex], but F(a) is a constant, so that's 0.
 

matt grime

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the derivative of F(a) (for a a constant) is not the same as F'(a).
 

quasar987

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ok sorry about the bad notation..
 

ryt

10
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thx :) i understand it now
 

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