Solve the Integral of x - 2|x| from [-1,2] | Easy Step-by-Step Solution

  • Thread starter Thread starter Shaybay92
  • Start date Start date
  • Tags Tags
    Integral
Shaybay92
Messages
122
Reaction score
0

Homework Statement



Integral[ x - 2|x|]dx from [-1,2]


The Attempt at a Solution



Wouldn't it become x^2/2 - 2x^2/2 so x^2/2 - x^2 and then -x^2/2?

So from [-1,2] it would just be

-((2^2)/2) - - ((-1)^2/2) which is -1.5 but the answer is -3.5...
 
Physics news on Phys.org
dddd
 
Wait, sorry this shouldn't be the issue because from my graph the whole function is negative anyway... any suggestions?
 
I can be wrong, but I would try

\int_{-1}^2f(x)dx = \int_{-1}^0f(x)dx + \int_{0}^2f(x)dx
 
I agree with Borek. The idea is that you can eliminate the absolute values by looking at the integrand on [-1, 0] and on [0, 2].

If x <= 0, |x| = -x.
If x >= 0, |x| = x.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Back
Top