Thread: Integrals.

1. Integrals.

For each of the functions, f, below

a. find F(x) = integral of f(t)dt from 0 to x on [0,3]

b. Is F continuous on [0,3]?

c. does F' = f on [0,3]?

here are the functions

1.) f(x) = 2 + x , when [0,1]
= 4 - x^2, when (1,3]

2.) f(x) = x, when [0,1]
= x -1 when (1,2]
= 0, when (2,3]

2. Re: Where do i begin?

take it step by step...first we know that the functions give are piecewise function thus they each need to be differentiated separately.
I'll show you how to do #1:

a. find F(x) = integral of f(t)dt from 0 to x on [0,3]
check if the functions are continuous so you plug in 1 in both equations
f(1)= 2+1 = 3
f(1)= 4-1^2= 3
b. hence the functions share the endpoint at x=1 so the function is continuous
a. Now you can integrate 2+x from 0 to 1 and you get 2x+1/2x^2= 2.5
integrate 4-x^2 from the t to 3 where t approaches 1 so

http://www.numberempire.com/equation.render?\lim_{t\rightarrow%201}\int_{t}^{3 }4-x^2%20dx

you solve the integral treating t as a normal integer

http://www.numberempire.com/equation.render?4x-\frac{{x}^{3}}{3}\mid%20[t,3]

http://www.numberempire.com/equation.render?\lim_{t\rightarrow%201}%204t-\frac{{t}^{3}}{3}-12+9
and finally you get 2/3 but u have to add the area from 0 to 1 so you add 5/2 to 2/3, and you get 19/6....i think this should be the right answer unless i messed up in the algebra. sorry about the links just follow them and you can see the equations!!

part C, you just have to find the value of the derivative at x=1 so just find the derivative normally for 2+x which is just 1, and then check if that equals the limit x-->1 of 2x and you get 2 so no the F' does not equal f

3. Re: Where do i begin?

To examine if the function f is continuous in a certain point you can also use the fact if a function f is differentiable in a point a then f is automatically continuous in a.
The reverse (if f is continuous in a then f is automatically differentiable in a) doesn't count always!