# Define g(x) as a function of $x$

Let $f(x)= \begin{cases} -1 & ,-2\leq x\leq 0 \\ \\ |x-1| & ,0<x\leq2 \end{cases}$ and $g(x)=\int_{-2}^{x}f(t) dt$.Define g(x) as a function of $x$.

I tried to solve it.I redefined $f(x)= \begin{cases} -1 & ,-2\leq x\leq 0 \\ 1-x & ,0<x<1 \\ x-1 & ,1\leq x\leq2 \end{cases}$

$g(x)=\int_{-2}^{x}f(t) dt=\begin{cases} \int_{-2}^{x}-1dx & ,-2\leq x\leq 0 \\ \int_{-2}^{x}(1-x)dx & ,0<x<1 \\ \int_{-2}^{x}(x-1)dx & ,1\leq x\leq2 \end{cases}$

$g(x)= \begin{cases} -(x+2) & ,-2\leq x\leq 0 \\ x-\frac{x^2}{2}+4 & ,0<x<1 \\ \frac{x^2}{2}-x-4 & ,1\leq x\leq2 \end{cases}$

But the answer is given to be
$g(x)= \begin{cases} -(x+2) & ,-2\leq x\leq 0 \\ x-\frac{x^2}{2}-2 & ,0<x<1 \\ \frac{x^2}{2}-x-1 & ,1\leq x\leq2 \end{cases}$

Have i done it wrong,what should be the correct method.Please guide me.

For $0<x<1$ we have \begin{align*} g(x)&=\int_{-2}^{x}f(t)dt\\ &=\int_{-2}^0f(t)dt+\int_0^xf(t)dt\\ &=\int_{-2}^0(-1)dt+\int_0^x(1-t)dt\\ &=(-1)(0+2)+x-\frac{1}{2}x^2-0\\ g(x)&=-\frac{1}{2}x^2+x-2,\qquad \text{ for }\;\;0<x<1 \end{align*}

On the other hand, for $1\le x\le 2$ it follows \begin{align*} g(x)&=\int_{-2}^{x}f(t)dt\\ &=\int_{-2}^0f(t)dt+\int_0^1f(t)dt+\int_1^xf(t)dt\\ &=\int_{-2}^0(-1)dt+\int_0^1(1-t)dt+\int_1^x(t-1)dt\\ &=(-1)(0+2)+1-\frac{1}{2}(1)^2-0+\frac{1}{2}x^2-x-\left[\frac{1}{2}-1\right]\\ g(x)&=\frac{1}{2}x^2-x-1,\qquad \text{ for }\;\;1\le x\le 2 \end{align*}

Thus \begin{equation*} \color{blue}{g(x)= \begin{cases} -(x+2)&-2\le x\le 0\\ -\frac{1}{2}x^2+x-2& 0<x<1\\ \frac{1}{2}x^2-x-1& 1\le x\le 2 \end{cases} } \end{equation*}

Note that

$$g(1) = \int_{-2}^1 f(x)\ dx = \int_{-2}^0 f(x)\ dx + \int_{0}^1 f(x)\ dx \neq \int_{-2}^1 (1-x)\ dx$$

In a slightly different approach from the direct, and brute force one, we recognize that inasmuch as $g$ is differentiable, it is, of course, continuous. Then, we have that

$$g(x)= \begin{cases} -x+C_1 ,&x<0\\\\ x-\frac12x^2+C_2 ,&0<x<1\\\\ \frac12x^2-x+C_3 ,&1<x \end{cases}$$

Now, exploiting the continuity of $g$ is as simple as One, Two, Three.

One, we enforce $g(-2)=0$, which reveals immediately that $C_1=-2$.

Two, we enforce continuity of $g$ at $x=0$. This shows that $C_2=C_1$ and thus $C_2=-2$ also.

Three, we enforce continuity of $g$ at $x=1$. This shows that $\frac12+C_2=-\frac12+C_3$, which implies that $C_3=-1$.

Putting it together gives

$$g(x)= \begin{cases} -(x+2) ,&x<0\\\\ -\frac12x^2+x-2 ,&0<x<1\\\\ \frac12x^2-x-1 ,&1<x \end{cases}$$

as expected!