CK asked in Science & MathematicsMathematics · 4 weeks ago

# Integral help.....?

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• 4 weeks ago

(Definitely graph all the functions to get a better feel for what's happening)

We have a variable of integration itself as a function of x, and it's also discontinuous. Might want to take it step by step. (no?)

Let's first compute the integral only on [0, 1].

Here, the variable of integration is simple: x - [_x_] = x - 0 = x and so we have just a "normal integral"

int[0 to 1] e^(2x - [_2x_]) dx

The integrand is discontinuous at x=0.5 so we split into the sum of integrals from 0 to 0.5, and from 0.5 to 1.

-------On [0, 0.5]:

2x - [_2x_] = 2x - 0 = 2x

int[0 to 0.5] e^(2x)dx

= (1/2)e^(2x) [0 to 0.5]

= (1/2)(e - 1)

-------On [0.5, 1]

2x - [_2x_] = 2x - 1

int[0.5 to 1] e^(2x - 1) dx

= 1/(2e) e^(2x) [0.5 to 1]

= 1/(2e)(e^2 - e)

= (1/2)(e - 1)

The sum is e - 1

But we already sniffed out x - [_x_] has a period of 1, and 2x - [_2x_] of 0.5, and 0.5 fits a whole number of times into 1 meaning the integral is the same on any interval of length 1.

Since we found it on [0, 1], just multiply with 10.

= 10(e - 1)

• 4 weeks ago

Bernie Sanders is my favourite hobbit