Definite Integration
Integral with sin²x and geometric function
MJAT_TS7_P2
Grade 12
Question:
**Paragraph II:** Let $f(x)=\sin^2 x$ on $[0,\pi/2]\to[0,1]$ and $g(x)=\sqrt{(\pi/2)x-x^2}$ on $[0,\pi/2]\to[0,\infty)$. The value of $\dfrac{\pi}{2}\displaystyle\int_0^{\pi/2}f(x)g(x)\,dx - \int_0^{\pi/2}g(x)\,dx$ is:
Step-by-Step Solution
Key Concept: $g(x)=\sqrt{(\pi/2)x-x^2}=\sqrt{x(\pi/2-x)}$. The expression $\frac{\pi}{2}\int fg\,dx-\int g\,dx=\int g(\frac{\pi}{2}f-1)dx=\int g(x)(\frac{\pi}{2}\sin^2x-1)dx$. By symmetry or specific computation.
Value $=\mathbf{0}$.
Correct Answer: 0