Evaluating the Definite Integral: $\int_{-1}^{2} (\pi^2 x \sin (\pi x))dx$

To evaluate the definite integral $\int_{-1}^{2} (\pi^2 x \sin (\pi x))dx$, we use the integration by parts method.

Integration by Parts Method

The integration by parts formula is $\int u \, dv = uv - \int v \, du$.

We choose:

• $u = \pi^2 x$
• $dv = \sin(\pi x) dx$

Calculating the differentials and integral:

• $du = \pi^2 dx$
• $v = \int \sin(\pi x) dx = -\frac{1}{\pi} \cos(\pi x)$

Applying the Formula

Substitute these into the formula:

$ \int \pi^2 x \sin (\pi x) dx = (\pi^2 x) \left(-\frac{1}{\pi} \cos(\pi x)\right) - \int \left(-\frac{1}{\pi} \cos(\pi x)\right) (\pi^2 dx) $

Simplify the expression:

$ = -\pi x \cos(\pi x) + \pi \int \cos(\pi x) dx $

Evaluate the remaining integral:

$ \int \cos(\pi x) dx = \frac{1}{\pi} \sin(\pi x) $

Combine the terms to find the antiderivative:

$ -\pi x \cos(\pi x) + \pi \left(\frac{1}{\pi} \sin(\pi x)\right) = -\pi x \cos(\pi x) + \sin(\pi x) $

Evaluating the Definite Integral

Evaluate the antiderivative $F(x) = -\pi x \cos(\pi x) + \sin(\pi x)$ at the limits of integration [-1, 2]:

$ \int_{-1}^{2} (\pi^2 x \sin (\pi x))dx = \left[ -\pi x \cos(\pi x) + \sin(\pi x) \right]_{-1}^{2} $

Value at the upper limit ($x=2$):

$ F(2) = -\pi (2) \cos(2\pi) + \sin(2\pi) = -2\pi (1) + 0 = -2\pi $

Value at the lower limit ($x=-1$):

$ F(-1) = -\pi (-1) \cos(-\pi) + \sin(-\pi) = \pi (-1) + 0 = -\pi $

Calculate the definite integral:

$ F(2) - F(-1) = (-2\pi) - (-\pi) = -2\pi + \pi = -\pi $

The result of the integral is $-\pi$.