Integral Calculus
Area with GIF-transformed curve
MJMT_Full_Test_06
Grade 12
Question:
The area of the region bounded by $y=x^2$ and $y=\sec^{-1}[-\sin^2 x]$, where $[\cdot]$ is the GIF, is
$\dfrac{4\pi\sqrt{\pi}}{3}$
$\dfrac{2\pi\sqrt{\pi}}{3}$
$\dfrac{\pi\sqrt{\pi}}{3}$
$\dfrac{4\sqrt{\pi}}{3}$
Step-by-Step Solution
Key Concept: Since $\sin^2 x\in[0,1]$, we have $-\sin^2 x\in[-1,0]$. The GIF $[-\sin^2 x]=-1$ for most $x$, and $\sec^{-1}(-1)=\pi$. So the curve $y=\pi$ (horizontal line).
$y=\pi$ for all $x$ (where defined). Area $=\frac{4\pi\sqrt\pi}{3}$.
Correct Answer: 1