Question

Find the value of \(\displaystyle\int_{-1}^{3/2}|x\sin\pi x|\,dx\).

Hint:

Check the sign of x sin(pi x) separately on (-1,1) and (1,3/2) before removing the modulus.

Answer 1

Step 1: Understanding the Concept:
To integrate \(|x\sin\pi x|\), first find where \(x\sin\pi x\) changes sign on \([-1,3/2]\).
Simply integrating \(x\sin\pi x\) directly would give a wrong answer wherever the function is negative, since the modulus flips its sign there.

Step 2: Finding the sign of \(x\sin\pi x\) on each part:
\(\sin\pi x=0\) at \(x=-1,0,1\) inside this interval.
On \((-1,0)\): \(x<0\) and \(\sin\pi x<0\) (since \(\pi x\in(-\pi,0)\)), so the product is positive.
On \((0,1)\): \(x>0\) and \(\sin\pi x>0\), so the product is positive.
On \((1,3/2)\): \(x>0\) but \(\sin\pi x<0\) (since \(\pi x\in(\pi,3\pi/2)\)), so the product is negative.
So \(|x\sin\pi x|=x\sin\pi x\) on \([-1,1]\) and \(|x\sin\pi x|=-x\sin\pi x\) on \([1,3/2]\).

Step 3: Splitting and evaluating the integral:
\[ I=\int_{-1}^{1}x\sin\pi x\,dx+\int_{1}^{3/2}(-x\sin\pi x)\,dx \]
Using integration by parts, \(\int x\sin\pi x\,dx=-\dfrac{x\cos\pi x}{\pi}+\dfrac{\sin\pi x}{\pi^2}+C\).
Let \(F(x)=-\dfrac{x\cos\pi x}{\pi}+\dfrac{\sin\pi x}{\pi^2}\).
\(F(1)=\dfrac{1}{\pi}\), \(F(-1)=-\dfrac{1}{\pi}\), \(F(3/2)=-\dfrac{1}{\pi^2}\).

Step 4: Substituting the limits:
\[ \int_{-1}^{1}x\sin\pi x\,dx=F(1)-F(-1)=\dfrac{1}{\pi}-\left(-\dfrac{1}{\pi}\right)=\dfrac{2}{\pi} \]
\[ \int_{1}^{3/2}x\sin\pi x\,dx=F(3/2)-F(1)=-\dfrac{1}{\pi^2}-\dfrac{1}{\pi} \]
So the second piece contributes \(\dfrac{1}{\pi^2}+\dfrac{1}{\pi}\) after reversing sign.
\[ I=\dfrac{2}{\pi}+\dfrac{1}{\pi}+\dfrac{1}{\pi^2}=\dfrac{3}{\pi}+\dfrac{1}{\pi^2}=\dfrac{3\pi+1}{\pi^2} \]

Final Answer:
The value of the definite integral after correctly handling the modulus is:
\[ \boxed{I=\dfrac{3\pi+1}{\pi^2}} \]