- Thread starter
- #1
paulmdrdo
Active member
- May 13, 2013
- 386
$\displaystyle\int|2x-1|dx$
please tell me what is the first step to solve this.
please tell me what is the first step to solve this.
There is only one definition: a piecewise function with two subdomains. Integrate at each subdomain.does that mean i have to integrate those two definitions?
Check the first one.$\displaystyle \int (2x-1)dx = {x^2}+x+C\,\,if\,x\geq\frac{1}{2}$
$\displaystyle \int (1-2x)dx = x-{x^2}+C\,\,if\,x\leq\frac{1}{2}$
Besides $x^2-x$ in the first formula, an indefinite integral, by definition, is a differentiable, and therefore a continuous, function. So, the constants in the two lines have to be such that at 1/2 the functions have the same value.we will have two answers is this correct?
$\displaystyle \int (2x-1)dx = {x^2}+x+C\,\,if\,x\geq\frac{1}{2}$
$\displaystyle \int (1-2x)dx = x-{x^2}+C\,\,if\,x\leq\frac{1}{2}$
You still compute two definite integrals: from -1 to 1/2 plus from 1/2 to 1.follow-up question, what if it is bounded?
$\displaystyle\int_{-1}^1|2x-1|dx$
follow-up question, what if it is bounded?
$\displaystyle\int_{-1}^1|2x-1|dx$
Look at the plot here.why is it from-1 to 1/2 plus 1/2 to 1?
It's not like there are many lines with formulas and constants in the first quote in post #8. Anyway, to understand this I suggest you write explicitly the answer to your original problem in post #1, i.e., the result of evaluating this indefinite integral. And remember that the resulting function should be differentiable and, in particular, continuous. .evgenymakarov "So, the constants in the two lines have to be such that at 1/2 the functions have the same value." - what do you mean by this?
and why is it from-1 to 1/2 plus 1/2 to 1?
Right. And considering it, we can express the solution in a more elegant way:Besides $x^2-x$ in the first formula, an indefinite integral, by definition, is a differentiable, and therefore a continuous, function. So, the constants in the two lines have to be such that at 1/2 the functions have the same value.
I assume this is related to evaluating $\int_{-1}^1|2x-1|\,dx$, but I am not sure how it is related.what i have in my mind is like this
$\displaystyle\int_{-1}^1 (2x-1)dx$
$\displaystyle\int_{-1}^1 (1-2x)dx$