Definite Integration
Product of sin²x and sqrt function
MJAT_TS7_P2
Grade 12
Question:
**Paragraph II (continued):** The value of $\dfrac{16}{\pi}\displaystyle\int_0^{\pi/2} f(x)g(x)\,dx$ is:
Step-by-Step Solution
Key Concept: $\int_0^{\pi/2}\sin^2 x\sqrt{\frac{\pi}{2}x-x^2}\,dx$. Use the substitution $x=\frac{\pi}{4}(1+\sin t)$ (since $g$ is a semicircle of sorts centered at $\pi/4$). The integral evaluates to $\pi/64$.
Value $=\mathbf{0.25}$.
Correct Answer: 0.25