JEE Advanced 2022 Paper 2 · Q03 · Definite Integration

Question

The greatest integer less than or equal to $\displaystyle\int_{1}^{2}\log_{2}(x^{3}+1)\,dx + \int_{1}^{\log_{2} 9}(2^{x}-1)^{1/3}\,dx$ is _____.

SolveKar AI · Accepted Answer

Substitute $u=\log_2(x^3+1)$ in the first integral; then in the second integral set $y=\log_2(x^3+1)$ so that the two integrals together form a complementary area. Specifically, the curves $y=\log_2(x^3+1)$ and $x=(2^y-1)^{1/3}$ are inverses, so the sum of the two integrals equals the area of the rectangle $[1,2]	imes[1,\log_2 9]$ adjusted: $\int_1^2\log_2(x^3+1)\,dx+\int_1^{\log_2 9}(2^y-1)^{1/3}\,dy = 2\log_2 9 - 1 = 2\log_2 9 - 1$. Numerically $2\log_2 9 - 1 \approx 2(3.1699)-1 = 5.3399$, so the greatest integer $\le$ this is $5$.

JEE Advanced 2022 Paper 2 · Q03 · Definite Integration

Solution