Definite integral question involving magnitude

Question

How to solve this using following method.

$\int_0^1 |x-5| dx$

I know how to use similar question.

If we have given -

$\int_1^5 |x-3| dx$

Then using method it changes into -

$\int_1^3 -(x-5) dx + \int_3^5 (x-5) dx$

Then we integrate.

But I don't know how to change above question using this method.

Answer 1

For $0 < x < 1$, we know that $x-5 <0$, and thus, we have $|x-5| = -(x-5) = 5-x$ using the definition of the modulus operator.

Now the problem reduces to $\int_{0}^{1} \displaystyle (5-x)\mathrm{d}x$. Hope it helps.