Q.Evaluate $\int_{-1}^{3/2} \left| x \sin(\pi x) \right|\, dx$

The absolute value forces a split wherever $x\sin(\pi x)$ changes sign on $[-1,\tfrac32]$.

Sign of $x\sin(\pi x)$:

• On $(-1,0)$: $x<0$ and $\sin(\pi x)<0$, so the product is positive.
• On $(0,1)$: $x>0$ and $\sin(\pi x)>0$, so positive.
• On $(1,\tfrac32)$: $x>0$ and $\sin(\pi x)<0$, so negative.

Hence $|x\sin\pi x|=x\sin\pi x$ on $[-1,1]$ and $=-x\sin\pi x$ on $[1,\tfrac32]$.

Antiderivative (by parts): with $u=x,\ dv=\sin(\pi x)\,dx$,

Values: $G(-1)=-\tfrac1\pi,\ G(0)=0,\ G(1)=\tfrac1\pi,\ G(\tfrac32)=-\tfrac1{\pi^2}.$

Pieces: …

Intuition

An absolute value can never be integrated with one formula across a sign change — $|f|$ equals $f$ where $f\ge 0$ and $-f$ where $f\le 0$. So the first job is to track the sign of $x\sin(\pi x)$ across $[-1,\tfrac32]$.

Step 1 — Sign analysis

Look at the two factors on each subinterval (note $\sin(\pi x)=0$ at the integers $x=-1,0,1$):

• $(-1,0)$: $x<0$; and $\pi x\in(-\pi,0)$ so $\sin(\pi x)<0$. Negative $\times$ negative $=$ positive.
• $(0,1)$: $x>0$; and $\pi x\in(0,\pi)$ so $\sin(\pi x)>0$. Positive.
• $(1,\tfrac32)$: $x>0$; and $\pi x\in(\pi,\tfrac{3\pi}{2})$ so $\sin(\pi x)<0$. Negative.

Therefore

Step 2 — An antiderivative of $x\sin(\pi x)$

Integrate by parts with $u=x$ (so $du=dx$) and $dv=\sin(\pi x)\,dx$ (so $v=-\tfrac{\cos\pi x}{\pi}$):

Evaluate at the break points (using $\cos(-\pi)=\cos\pi=-1$, $\cos\tfrac{3\pi}{2}=0$, $\sin\tfrac{3\pi}{2}=-1$):

$G(-1)=-\frac{(-1)(-1)}{\pi}+0=-\frac1\pi,\quad G(0)=0,$ …

Method: Integrating an absolute value — split at the sign changes

Use this for any $\int_a^b |f(x)|\,dx$. An absolute value has no single antiderivative across a sign change, so you must break the interval where $f$ changes sign and integrate each piece with the correct sign.

Steps

Step 1: Find where the inside changes sign.

Solve $f(x)=0$ inside $[a,b]$ and determine the sign of $f$ on each resulting subinterval (test a point, or reason factor-by-factor — e.g. for $x\sin(\pi x)$ track the signs of $x$ and of $\sin(\pi x)$ separately).

Step 2: Rewrite $|f|$ piecewise.

On subintervals where $f\ge 0$, $|f|=f$; where $f\le 0$, $|f|=-f$. This converts the modulus into ordinary signed integrals.

Step 3: Find one antiderivative of $f$ (here by parts). …

Common Mistakes

Mistake 1: Integrating $|x\sin\pi x|$ as if it were $x\sin\pi x$ over the whole interval.

Why it's wrong: the product changes sign at $x=1$ inside $[-1,\tfrac32]$, so a single antiderivative undercounts the area (the negative part subtracts instead of adding). Correct approach: split at every sign change and flip the sign where the inside is negative.

Mistake 2: Getting the sign wrong on $(-1,0)$.

Why it's wrong: there $x<0$ and $\sin(\pi x)<0$, so the product is positive (negative times negative); assuming it is negative flips a piece. Correct approach: check the sign of both factors on each subinterval. …