2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ lower bound is a negative one. $\endgroup$ – James Dec 26 '14 at 22:16
  • $\begingroup$ SORRY, problem was written incorrectly. $\endgroup$ – James Dec 28 '14 at 5:31
7
$\begingroup$

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:)

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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?

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ you want the lower limit $-1?$ $\endgroup$ – abel Dec 26 '14 at 23:50
  • $\begingroup$ @abel Yes, thanks. Typo fixed. $\endgroup$ – Ahaan S. Rungta Dec 27 '14 at 0:41
0
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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?

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.