I have trouble splitting the integral $$\int_{-1}^{6} (5-|2-x|)\, dx$$
Tried so far:
Split the 3 and the absolute value to two separate integrals. Draw absolute value graph. Integrate both. I think algebra may be the problem.
[Math] Find integral of absolute values by splitting integrals, $\int_{-1}^{4} (3-|2-x|)\, dx$
absolute valuecalculusdefinite integralsintegration
Related Solutions
$$I=\int_{-2\pi}^{2\pi} xe^{-\mid x\mid}dx.$$
Loosely speaking, you can think about the definite integral as the area bounded by the function $xe^{-\mid x\mid}$ and the $x$-axis, as the variable $x$ moves from $x=-2\pi$ to $x=0$ then from $x=0$ through to $x=2\pi$. So, intuitively it's not too much of a step to see that $$I=\int_{-2\pi}^{0} xe^{-\mid x\mid}dx+\int_{0}^{2\pi} xe^{-\mid x\mid}dx.$$ Notice in the left integral the $x$ values are only ever negative or zero, and in the right integral the $x$ values are only ever positive or zero, so we can rewrite the whole expression $$I=\int_{-2\pi}^{0} xe^{x}dx+\int_{0}^{2\pi} xe^{-x}dx,$$ since $-|x|=x$ for $x\leq 0$ and $-|x|=-x$ for $x\geq 0$. You can now evaluate the integrals separately to obtain the correct result. Hope this helps.
There are three regions:
$$\int_{-1}^{2} (|x|+|1-x|) dx=\int_{-1}^{0}-2x+1 \ dx +\int_{0}^{1}1\ dx+\int_{1}^{2}2x-1 \ dx=5.$$
To verify this, take a look at the following figure:
Best Answer
Hint. You may write $$ \begin{align} \int_{-1}^4 (3-|2-x|)\, dx &=\int_{-1}^2 (3-|2-x|)\, dx+\int_2^4 (3-|2-x|)\, dx\\\\ &=\int_{-1}^2 (3-(2-x))\, dx+\int_2^4 (3-(-(2-x)))\, dx \\\\ &=\int_{-1}^2 (1+x)\, dx+\int_2^4 (5-x)\, dx \\\\ &= ... \end{align} $$ where we have used $$ |u| = \begin{cases} -u, & \text{if $u \leq 0$} \\ u, & \text{if $u \geq 0$.} \end{cases} $$ I think you can take it from here.
Addendum: I've change the lower bound, as you changed it:)