JEE Advanced 2026 Paper 2, Mathematics Section 4 Q2: Area as Definite Integral

From the previous part, ${\alpha }_1=0$, ${\alpha }_2=\frac{\pi }{2}$, ${\alpha }_3=2\pi$, ${\alpha }_4=\frac{5\pi }{2}$.

$\beta ={\int }_0^{5\pi /2}\left|e^{-x}(\sin x+\cos x)-e^{-x}\right|dx$.

The sign of $\sin x+\cos x-1$ changes at the ${\alpha }_i$. Standard sign analysis on the intervals $\left[0,\frac{\pi }{2}\right]$, $\left[\frac{\pi }{2},2\pi \right]$, $\left[2\pi ,\frac{5\pi }{2}\right]$ gives:

$\beta ={\int }_0^{\pi /2}\left(e^{-x}(\sin x+\cos x)-e^{-x}\right)dx+{\int }_{\pi /2}^{2\pi }\left(e^{-x}-e^{-x}(\sin x+\cos x)\right)dx+{\int }_{2\pi }^{5\pi /2}\left(e^{-x}(\sin x+\cos x)-e^{-x}\right)dx$.

Using the antiderivatives $\int e^{-x}\sin xdx=-\frac{e^{-x}(\sin x+\cos x)}{2}$ and $\int e^{-x}\cos xdx=\frac{e^{-x}(\sin x-\cos x)}{2}$, an antiderivative of $e^{-x}(\sin x+\cos x)$ is $-e^{-x}\sin x$ (verify by differentiating), and an antiderivative of $e^{-x}$ is $-e^{-x}$.

So $F(x)=-e^{-x}\sin x+e^{-x}$ is an antiderivative of $e^{-x}(\sin x+\cos x)-e^{-x}$. Evaluating at the relevant endpoints and combining with the sign-flips gives

$\beta =2e^{-\pi /2}+e^{-5\pi /2}$.

Therefore $\beta -2e^{-\pi /2}=e^{-5\pi /2}$, and

$-\frac{1}{\pi }{\log }_e\left(\beta -2e^{-\pi /2}\right)=-\frac{1}{\pi }\cdot \left(-\frac{5\pi }{2}\right)=\frac{5}{2}=2.50$.