Animated Solution for Mathematics - Definite Integration: The integral ${\int }_{-1}^{3/2}(|{\pi }^{2}x\sin (\pi x)|)dx$ is equal to:

Welcome, future engineers! Today, we are going to dissect a problem that often trips up even the most prepared students. We are looking at the integral $I={\int }_{-1}^{3/2}|{\pi }^{2}x\sin (\pi x)|dx$.

At first glance, the presence of the modulus sign and the trigonometric product might seem daunting. But remember, in the world of JEE Advanced, complexity is often just a mask for a beautiful, structured reality. Let us peel back that mask.

The Modulus Trap

First, let us simplify our life. The term ${\pi }^2$ is a positive constant. We can pull it out of the modulus and the integral entirely, leaving us with:

Now, the core of the problem lies in the behavior of $f(x)=x\sin (\pi x)$. To handle the modulus, we must know where this function is positive and where it is negative.

The roots of $x\sin (\pi x)=0$ are the keys to the kingdom. Within our interval $[-1,3/2]$, the roots are $x=-1,0,1$. These are the points where our curve kisses the x-axis.

The Art of Splitting

Imagine the graph. From $x=-1$ to $x=1$, the product $x\sin (\pi x)$ remains non-negative.

In the interval $[-1,0]$, both $x$ and $\sin (\pi x)$ are negative, so their product is positive. In $[0,1]$, both are positive. However, once we cross $x=1$, $x$ remains positive, but $\sin (\pi x)$ dips into negative territory.

This is where the modulus demands our attention. We must split the integral at $x=1$:

That minus sign on the second integral is crucial—it flips the negative area into a positive one, satisfying the definition of the absolute value.

The Integration Engine

Now, we need the antiderivative of $x\sin (\pi x)$. This is a classic application of Integration by Parts.

Using the ILATE rule, we set $u=x$ and $dv=\sin (\pi x)dx$. This gives us $du=dx$ and $v=-\frac{\cos (\pi x)}{\pi }$. Applying the formula $\int udv=uv-\int vdu$, we get:

This expression is our workhorse. We will use it to evaluate both parts of our split integral.

The Final Dance

Let us compute the first part, $I_1={\int }_{-1}^{1}x\sin (\pi x)dx$. Evaluating our antiderivative at the limits $1$ and $-1$, we find $I_1=\frac{2}{\pi }$.

Next, for the second part, $I_2={\int }_1^{3/2}x\sin (\pi x)dx$, we evaluate at $3/2$ and $1$, yielding $I_2=-\frac{1}{{\pi }^2}-\frac{1}{\pi }$.

Finally, we combine them: $I={\pi }^2[I_1-I_2]$. Substituting our values:

Distributing the ${\pi }^2$ and simplifying, the terms collapse beautifully into the final result:

There you have it! What seemed like a terrifying integral was just a matter of understanding the geometry of the function and applying the right tools with precision. Keep practicing, stay curious, and remember: every problem is just a story waiting to be solved.