f(x) = summation from n equals 0 to infinity of the quotient of the quantity n plus 1 and 3 to the n plus 1 power times x to the nth power with an interval of convergence, –3 < x < 3. Find exactly the value of the integral from 0 to 2 of of x, dx. Your answer will be a positive integer. Type your answer in the space below (ex. 4).
thank you so much for your help!
We can find the integral from 0 to 2 of x, dx by antiderivative and applying the Fundamental Theorem of Calculus.
Let's begin by finding the antiderivative of x with respect to x.
∫ x dx = (1/2)x^2 + C
Now, we can evaluate the definite integral from 0 to 2.
∫[0 to 2] x dx = [(1/2)x^2] from 0 to 2
= [(1/2)*(2)^2] - [(1/2)*(0)^2]
= (1/2)*(4) - (1/2)*(0)
= 2
Therefore, the value of the integral from 0 to 2 of x, dx is 2.
To find the value of the integral from 0 to 2 of x, we can integrate the function f(x) = x over the given interval, which is -3 < x < 3.
Step 1: Evaluate the indefinite integral
The indefinite integral of x with respect to x is given by:
∫ x dx = (1/2) * x^2 + C,
where C is the constant of integration.
Step 2: Evaluate the definite integral
To find the definite integral from 0 to 2, we substitute the upper and lower limits into the indefinite integral:
∫[0 to 2] x dx = [(1/2) * x^2] [0 to 2]
= (1/2) * (2^2) - (1/2) * (0^2)
= (1/2) * 4 - (1/2) * 0
= 2 - 0
= 2.
Therefore, the value of the integral from 0 to 2 of x is 2, which is a positive integer.
I think this converges only for x=0. Consider what happens when x is any small number, say 0.01
Then you have then nth term greater than 1 when
(n/3)^n * 0/01^(n-1) = (0.00333333n)^n >1
clearly that holds when 0.00333333n > 1
So, for any |x| > 0 there will come a point where the nth term exceeds 1 for sufficiently large n.