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I am trying to evaluate the integral $\int_{-1}^2 (x-2|x|)\,dx$

I know that this should give me $x^2 /2 - x^2$ for the antiderivative.

I then evaluate at $2$ which gives me $2 - 4 = -2$

Then evaluate at $-1$ and get $\frac{1}{2} - 1 = -\frac{1}{2}$.

Then I find the difference $-2 - \frac{1}{2} = -2.5$

This is wrong and I do not know why.

2 Answers 2

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You should consider splitting the integral at the point where $|x|$ changes from $-x$ to $+x$, namely at $x=0$. So $ \int_{-1}^2(x-2|x|)dx = \int_{-1}^0(x-2(-x))dx + \int_0^{2}(x-2x)dx $ $ =3\int_{-1}^0 xdx -\int_0^2xdx $

which are integrals you should be able to evaluate.

  • 0
    In the range $-1 \leq x \leq 0$ we have that $|x| = -x$, so I substitute $(-x)$ where $|x|$ was in the first integral. Then $x-2(-x) = 3x$ which is where $3\int_{-1}^0xdx$ comes from. Similarly, in the range $0 \leq x \leq 2$ we have that $|x| = x$, so I replace $|x|$ by $x$ in the second integral. Then $x-2x = -x$, which is where $-\int_0^2xdx$ comes from.2012-04-22
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Hint

$|x|=\begin{cases} -x, &x \le 0\\x. &x \ge 0\end{cases}$

So, $\begin{align}\int_{-1}^2|x|\; \mathrm{d} x&=\int_{-1}^0|x| \;\mathrm{d}x+\int_0^2|x| \;\mathrm{d}x\\&=?\end{align}$

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    |x| will be positive, yes, but the x-coordinate is negative meaning it's relation to$x$is -x.2016-03-08