Question

Evaluate:
$\displaystyle \int_{0}^{3} x \cos(\pi x) \, dx$

Answer 1

To evaluate:

$\displaystyle \int_{0}^{3} x \cos(\pi x) \, dx$

Step 1: Use integration by parts

Let:

• $u = x \Rightarrow du = dx$
• $dv = \cos(\pi x) \, dx \Rightarrow v = \dfrac{1}{\pi} \sin(\pi x)$

Using the formula:

$\displaystyle \int u \, dv = uv - \int v \, du$

Apply the limits:

$\displaystyle \int_{0}^{3} x \cos(\pi x) \, dx = \left[ x \cdot \dfrac{1}{\pi} \sin(\pi x) \right]_0^3 - \int_{0}^{3} \dfrac{1}{\pi} \sin(\pi x) \, dx$

Evaluate the first term:

$\left[ \dfrac{x \sin(\pi x)}{\pi} \right]_0^3 = \dfrac{3 \sin(3\pi)}{\pi} - \dfrac{0 \cdot \sin(0)}{\pi} = \dfrac{3 \cdot 0}{\pi} = \mathbf{0}$

Evaluate the second term:

$\displaystyle \int_{0}^{3} \dfrac{1}{\pi} \sin(\pi x) \, dx = \dfrac{1}{\pi} \left[ -\dfrac{1}{\pi} \cos(\pi x) \right]_0^3 = -\dfrac{1}{\pi^2} \left[ \cos(3\pi) - \cos(0) \right]$

$= -\dfrac{1}{\pi^2} \left[ -1 - 1 \right] = -\dfrac{1}{\pi^2} (-2) = \mathbf{\dfrac{2}{\pi^2}}$

Final Answer:

$\displaystyle \int_{0}^{3} x \cos(\pi x) \, dx = \mathbf{0 + \dfrac{2}{\pi^2} = \dfrac{2}{\pi^2}}$