To evaluate the integral
\[ I = \int_0^\pi x^2 \sin(x) , dx, \]
we can use integration by parts. Let's recall the formula for integration by parts:
\[ \int u , dv = uv - \int v , du, \]
where we choose \( u \) and \( dv \) appropriately.
We can make the following choices:
- Let \( u = x^2 \) which gives us \( du = 2x , dx \).
- Let \( dv = \sin(x) , dx \) which gives us \( v = -\cos(x) \).
Now, applying integration by parts, we have:
\[ I = \left[ -x^2 \cos(x) \right]_0^\pi - \int_0^\pi -\cos(x)(2x) , dx. \]
Now, we evaluate the boundary term:
\[ \left[ -x^2 \cos(x) \right]_0^\pi = -\left( \pi^2 \cos(\pi) - 0^2 \cos(0) \right) = -\left( \pi^2 (-1) - 0 \right) = \pi^2. \]
So, the integral turns into:
\[ I = \pi^2 + 2 \int_0^\pi x \cos(x) , dx. \]
Next, we need to evaluate the integral \( \int_0^\pi x \cos(x) , dx \) again using integration by parts. We can choose:
- \( u = x \) which gives \( du = dx \),
- \( dv = \cos(x) , dx \) which results in \( v = \sin(x) \).
Applying integration by parts:
\[ \int x \cos(x) , dx = \left[ x \sin(x) \right]_0^\pi - \int_0^\pi \sin(x) , dx. \]
Now we evaluate the boundary term:
\[ \left[ x \sin(x) \right]_0^\pi = \pi \sin(\pi) - 0 \sin(0) = \pi \cdot 0 - 0 = 0. \]
This integral simplifies to:
\[ \int_0^\pi x \cos(x) , dx = 0 - \left[ -\cos(x) \right]_0^\pi = 0 - (-\cos(\pi) + \cos(0)) = 0 - (-(-1) + 1) = 0 - (1 + 1) = 0. \]
Thus,
\[ \int_0^\pi x \cos(x) , dx = 0. \]
Now substituting this back into our expression for \( I \):
\[ I = \pi^2 + 2 \cdot 0 = \pi^2. \]
Finally, we conclude that the value of the integral is
\[ \int_0^\pi x^2 \sin(x) , dx = \pi^2. \]