Question

how do i intergrate -2 lxl +1

Answer 1

\[
-2|x|+1 = \begin{cases}
-2x+1&x>0\\
2x+1&x<0
\end{cases}
\]

Answer 2

\[
\int _a^b-2|x|+1\;dx = \int_{a}^02x+1\;dx+\int_0^b-2x+1\;dx
\]

Answer 3

The antiderivative of \(|x|\) would be: \[
\begin{split}
\int
\begin{cases}
x&x>0\\
-x&x<0
\end{cases}
dx
&=
\begin{cases}
\frac{x^2}{2}+C&x>0\\
-\frac{x^2}{2}+C&x<0
\end{cases}\\
&=
C+\frac{x}{2}\cdot \begin{cases}
x&x>0\\
-x&x<0
\end{cases}\\
&= \frac{x|x|}{2}+C
\end{split}
\]

Answer 4

Using properties of integrals, we would say \[
\int -2|x|+1\;dx =\color{red}{ -2\int |x|\;dx}+\color{blue}{\int 1\;dx }= \color{red}{-x|x|} +\color{blue}{ x} +C
\]

Answer 5

if im calculating a definite integral from -1 to 1, I got my answer to be 2. would that be right?