Question

$\displaystyle\int_{0}^{1} x\sin(\pi x)\,dx$ is equal to

• A. 1
• B. $1/2$
• C. $\pi$
• D. $1/\pi$

Hint:

Integration by parts: choose $u$ as algebraic and $dv$ as trigonometric. Evaluate the boundary terms carefully noting $\cos(0)=1$, $\cos(\pi)=-1$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Use integration by parts: $\int u\,dv = uv - \int v\,du$.
Step 2: Detailed Explanation:
Let $u=x$, $dv=\sin(\pi x)dx$. Then $du=dx$, $v=-\dfrac{\cos\pi x}{\pi}$.
$\left[-\dfrac{x\cos\pi x}{\pi}\right]_0^1 + \int_0^1 \dfrac{\cos\pi x}{\pi}dx = -\dfrac{1\cdot(-1)}{\pi} + \left[\dfrac{\sin\pi x}{\pi^2}\right]_0^1 = \dfrac{1}{\pi} + 0 = \dfrac{1}{\pi}$.
Step 3: Final Answer:
The integral equals $\dfrac{1}{\pi}$.