Animated Solution for Mathematics - Definite Integration: Find the value of ${\int }_{-1}^{3/2}|x\sin \pi x|dx$.

Welcome, fellow traveler of the JEE Advanced path. Today, we are not just solving an integral; we are dissecting a beautiful, symmetric, and slightly deceptive function:

When you see a modulus sign in an integral, do not panic. Think of it as a challenge to your geometric intuition. The modulus is a filter—it demands that every piece of area we calculate be positive. Our job is to find where the function dips below the x-axis and 'flip' it back up.

Finding the Hinge

Before we touch our pens to paper, we must visualize the terrain. We are integrating from $x=-1$ to $x=1.5$. The function inside is $f(x)=x\sin \pi x$.

Where does this function cross the x-axis? We set $x\sin \pi x=0$. This happens when $x=0$ or when $\sin \pi x=0$. The sine function hits zero at integer multiples of $\pi$. Thus, $\pi x=n\pi$, which means $x$ must be an integer.

Within our interval $[-1,1.5]$, the critical crossing point is $x=1$. This is our 'hinge.' We cannot integrate across this point blindly; we must split our journey into two distinct paths: from $-1$ to $1$, and from $1$ to $1.5$.

The Gift of Symmetry

Let us look at the first leg of our journey: ${\int }_{-1}^1|x\sin \pi x|dx$. Here, the limits are symmetric. Whenever you see limits like $[-a,a]$, your first instinct should be to check for symmetry.

Let us test $f(x)=x\sin \pi x$. Replacing $x$ with $-x$, we get $f(-x)=(-x)\sin (\pi (-x))$. Since $\sin (-\theta )=-\sin (\theta )$, this becomes $(-x)(-\sin \pi x)=x\sin \pi x$. It is an even function!

This is a massive gift. It means the area from $-1$ to $0$ is identical to the area from $0$ to $1$. Instead of doing the work twice, we can simply calculate ${\int }_0^{1}x\sin \pi xdx$ and multiply the result by $2$.

Furthermore, in the interval $[0,1]$, both $x$ and $\sin \pi x$ are positive, so we can drop the modulus bars entirely. The integral becomes:

The Heavy Lifting (Integration by Parts)

Now, we face the core of the problem: $\int x\sin \pi xdx$. We have a product of an algebraic function ($x$) and a trigonometric function ($\sin \pi x$). This is the classic setup for Integration by Parts (IBP). Recall the formula: $\int udv=uv-\int vdu$.

Following the ILATE rule, we choose $u=x$ (Algebraic) and $dv=\sin \pi xdx$ (Trigonometric). Calculating the components:

- $u=x⟹du=dx$ - $dv=\sin \pi xdx⟹v=-\frac{\cos \pi x}{\pi }$

Applying the formula, we get:

Simplifying this, we arrive at the antiderivative:

Evaluating this from $0$ to $1$: - At $x=1$: $-\frac{1\cdot \cos \pi }{\pi }+\frac{\sin \pi }{{\pi }^2}=-\frac{-1}{\pi }+0=\frac{1}{\pi }$. - At $x=0$: The expression is $0$.

So, the first part of our integral is $2\times \frac{1}{\pi }=\frac{2}{\pi }$.

The Final Stretch

We are not done yet! We still have the second interval: ${\int }_1^{3/2}|x\sin \pi x|dx$. In this region, the angle $\pi x$ ranges from $\pi$ to $\frac{3\pi }{2}$, which is the third quadrant.

In the third quadrant, sine is negative. Since $x$ is positive, the product $x\sin \pi x$ is negative. To keep the area positive, we must integrate the negative of the function: ${\int }_1^{3/2}-x\sin \pi xdx$.

Using the same antiderivative we found earlier, we evaluate:

Plugging in the upper limit $x=\frac{3}{2}$:

Plugging in the lower limit $x=1$:

Subtracting the lower from the upper and applying the outer negative sign:

Finally, we sum our two parts:

Combining these into a single fraction, we get:

Take a moment to appreciate this. We navigated the modulus, utilized symmetry, executed integration by parts, and carefully managed the signs in the third quadrant. This is the essence of JEE Advanced mathematics—a symphony of logical steps leading to a precise, elegant conclusion. Well done!