Definite Integration
Integrals involving 1/x − ln(x) and matching sums
MJAT_TS4_P1
Grade 12
Question:
Let $\displaystyle\sum_{r=1}^\infty\frac{1}{r^2}=a$, $f(x)=\dfrac{1-\ln x}{x}$, $g(x)=\dfrac{1-\ln x}{x^2}$, $I_1=\displaystyle\int_0^1 f(x)\,dx$, $I_2=\displaystyle\int_0^1 g(x)\,dx$.
Match each entry in List-I to the correct entry in List-II.
**List-I:** P) $I_1$; Q) $I_2$; R) $I_1-I_2$; S) $\displaystyle\lim_{n\to\infty}\sum_{r=1}^n\frac{r(\ln r-\ln n)}{n^2-r^2}$
**List-II:** 1) $a-$?; 2) $-\frac{3}{4}a$; 3) $-a$; 4) $a+$?; 5) $a-2$
A) P→4, Q→2, R→5, S→1
B) P→5, Q→4, R→3, S→1
C) P→1, Q→3, R→2, S→3
D) P→1, Q→2, R→2, S→1
Step-by-Step Solution
Key Concept: $I_1=\int_0^1\frac{1-\ln x}{x}dx$: note $\int_0^1 x^{n-1}\ln x\,dx = -\frac{1}{n^2}$. So $I_1=\int_0^1\sum_{n=1}^\infty x^{n-1}(1-\ln x)dx = \sum\left(\frac{1}{n}+\frac{1}{n^2}\right)$... need care with convergence.
Answer: **C**. P→(1), Q→(3), R→(2), S→(3).
Correct Answer: C