$f(x)=\int_{-1}^{x}(1-\left | t \right |)dt(x\geq -1)$,The area of the curve $f(X)$ and x-axis

Question

$$ f(x)=\int_{-1}^{x}(1-\left | t \right |)dt(x\geq -1)$$ ,The area of the curve $f(X)$ and x-axis When I did the calculations, I used wolfram alpha.It is the use of a digital $1 + \sqrt2 $Look at this question, I do not know how this number is how come! I have tried wolframOh, it doesn't work. This is what I do, access may be a little slow.

$ f(x)=\int_{-1}^{x}(1-\left | t \right |)dt = -\dfrac{x\left|x\right|-2x-1}{2} $ \begin{cases} x-\dfrac{x^2}{2}+\dfrac{1}{2} , & \text{if $x>0$ } \\ x+\dfrac{x^2}{2}+\dfrac{1}{2}, & \text{if $x<0$ } \end{cases}

$f(x)=\int_{-1}^{0}(x+\dfrac{x^2}{2}+\dfrac{1}{2}) =\dfrac{1}{6}$

$f(x)=\int_{0}^{1+\sqrt{2}}(x+\dfrac{x^2}{2}+\dfrac{1}{2}) =\dfrac{5}{6}+\dfrac{2\sqrt{2}}{3}$

but I just can not remember why use $1+\sqrt{2}$
Reference answer is $1+\dfrac{2\sqrt{2}}{3}$

Answer 1

The upper boundary $1+\sqrt 2$ comes from the solution of

$$x-\dfrac{x^2}{2}+\dfrac{1}{2}=0$$

Because the area is between $f(x)$ and the $x$-axis and the upper integral boundary is the root of $f(x)=0$.