# Find integral of absolute values by splitting integrals, $\int_{-1}^{4} (3-|2-x|)\, dx$

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.

• lower bound is a negative one. – James Dec 26 '14 at 22:16
• SORRY, problem was written incorrectly. – James Dec 28 '14 at 5:31

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:)

Note that $$\displaystyle\int_{-1}^4 \left( 3 - |2-x| \right) \, \mathrm{d}x = \underbrace{\displaystyle\int_{-1}^4 3 \, \mathrm{d}x}_{=\, 9} - \displaystyle\int_{-1}^4 \left| 2 - x \right| \, \mathrm{d}x.$$I think this could be called splitting the integral, but what I think you mean is how you find $\displaystyle\int_{-1}^4 \left| 2-x \right| \, \mathrm{d}x$ by splitting. Really, you can find this just by graphing the function $y=|2-x|$ and finding areas. However, I think the following is what you are looking for.

Note that, if $x \le 2$, the integrand is $|2-x|=2-x$ and if $x \ge 2$, it is $x-2$. Hence, \begin {align*} \displaystyle\int_{-1}^4 \left|2-x\right| \, \mathrm{d}x &= \displaystyle\int_{-1}^{2} |2-x| \, \mathrm{d}x + \displaystyle\int_2^4 |2-x| \, \mathrm{d}x \\&= \displaystyle\int_{-1}^2 \left( 2 - x \right) \, \mathrm{d}x + \displaystyle\int_2^4 \left( x - 2 \right) \, \mathrm{d}x. \end {align*}Can you finish from here?

• you want the lower limit $-1?$ – abel Dec 26 '14 at 23:50
• @abel Yes, thanks. Typo fixed. – Ahaan S. Rungta Dec 27 '14 at 0:41

we have for $$x\geq 2$$ the integral $$\int_{2}^{4}5-xdx$$ and for $$x<2$$ the integral $$\int_{1}^{2}1+xdx$$

you could complete what you have started by splitting the integral into three pieces and interpreting each integral as the area of rectangles and triangles:

(a) A = $\int_{-1}^4 3\ dx,$

(b) B = $\int_{-1}^2 | x - 2| \ dx,$

(c) C = $\int_2^4 |x-2| \ dx$

$A = 15=$ the area of a rectangle of base $5$ and height 3.

$B = 9/2 =$ area of an isosceles right triangle of base 3.

$C = 2 =$ area of an isosceles right triangle of base 2.

putting all these together $$\int_{-1}^4 3 - |x-2| \ dx = 15 - 9/2 - 2 = 17/2$$

You could write like this$$\int_1^4(3-|2-x|)\ dx=\int_{-1}^43\ dx-\int_{-1}^4|2-x|\ dx.$$ Then note that $2-x\ge 0$ when $x\in[-1,2]$ and $2-x\le 0$ when $x\in[2,4]$. So we have $$\int_{-1}^43\ dx-\left(\int_{-1}^2(2-x)\ dx-\int_2^4(2-x)\ dx\right)$$ Can you take it from here?