Q.Evaluate $\int_{-1}^2 \left| x^3 - x \right|\, dx$

The key idea is that the absolute value forces us to split the integral at the points where $x^3 - x = 0$, i.e., where the expression changes sign.

Step 1: Find the roots.

$x^3 - x = x(x-1)(x+1) = 0$ gives $x = -1, 0, 1$. On $[-1, 2]$, the sign changes at $0$ and $1$.

Step 2: Determine the sign of $x^3 - x$ on each subinterval.

• On $(-1, 0)$: test $x = -0.5$ → $(-0.5)^3 - (-0.5) = -0.125 + 0.5 = 0.375 > 0$.
• On $(0, 1)$: test $x = 0.5$ → $0.125 - 0.5 = -0.375 < 0$.
• On $(1, 2)$: test $x = 1.5$ → $3.375 - 1.5 = 1.875 > 0$.

Thus $|x^3 - x| = x^3 - x$ on $[-1,0]$ and $[1,2]$, and equals $-(x^3 - x) = x - x^3$ on $[0,1]$.

Step 3: Write and evaluate the sum of integrals.

Compute each:

• From $-1$ to $0$: $\left[0\right] - \left[\frac{1}{4} - \frac{1}{2}\right] = 0 - (-\frac{1}{4}) = \frac{1}{4}$.
• From $1$ to $2$: $\left[\frac{16}{4} - \frac{4}{2}\right] - \left[\frac{1}{4} - \frac{1}{2}\right] = (4 - 2) - (-\frac{1}{4}) = 2 + \frac{1}{4} = \frac{9}{4}$.
• For $\int (x - x^3)\,dx = \frac{x^2}{2} - \frac{x^4}{4}$ from $0$ to $1$: $\left[\frac{1}{2} - \frac{1}{4}\right] - 0 = \frac{1}{4}$.

Sum: $\frac{1}{4} + \frac{1}{4} + \frac{9}{4} = \frac{11}{4}$.

The integrand $x^3-x=x(x-1)(x+1)$ has zeros at $x=-1,0,1$. Its sign on $[-1,2]$ is:

• $[-1,0]$: positive, so $|x^3-x| = x^3-x$;
• $[0,1]$: negative, so $|x^3-x| = -(x^3-x)$;
• $[1,2]$: positive, so $|x^3-x| = x^3-x$.

With $\displaystyle\int (x^3-x)\,dx = \frac{x^4}{4}-\frac{x^2}{2} = F(x)$:

Adding: $\displaystyle \frac14+\frac14+\frac94=\frac{11}{4}$.

Method: Splitting a Definite Integral of an Absolute Value

Use this when the integrand contains $|f(x)|$: break the interval at the points where $f$ changes sign, and drop the modulus with the correct sign on each piece.

Steps

Step 1: Find where $f(x)=0$ inside the interval.

Factor $f$ and locate its roots. For $|x^3-x|=|x(x-1)(x+1)|$, the roots are $x=-1,0,1$.

Step 2: Determine the sign of $f$ on each subinterval.

Test a point in each piece. On $[-1,0]$, $f>0$ so $|f|=f$; on $[0,1]$, $f<0$ so $|f|=-f$; on $[1,2]$, $f>0$ so $|f|=f$.

Step 3: Integrate each piece with its sign and add.

Compute $\int$ of the signed expression over each subinterval and sum the (non-negative) contributions:

Common Mistakes

Mistake 1: Integrating $|x^3-x|$ as $x^3-x$ over the whole interval.

Why it's wrong: ignoring the sign changes lets positive and negative areas cancel, giving too small a value. Correct approach: split at the roots and use $|f|$ correctly.

Mistake 2: Getting the sign of $f$ wrong on a subinterval.

Why it's wrong: on $[0,1]$, $x^3-x<0$, so $|f|=-(x^3-x)$; using $+f$ there flips a term. Correct approach: test the sign on each piece.

Mistake 3: Missing a root inside the interval.

Why it's wrong: overlooking $x=0$ merges two pieces of opposite sign. Correct approach: find all zeros of $f$ in $[a,b]$ before splitting.