Definite Integration Question 18
Question 18 - 31 January - Shift 1
Let $\alpha \in(0,1)$ and $\beta=\log _{e}(1-\alpha)$. Let
$P_n(x)=x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+\ldots .+\frac{x^{n}}{n}, x \in(0,1)$.
Then the integral $\int_0^{\alpha} \frac{t^{50}}{1-t} dt$ is equal to
(1) $\beta-P _{50}(\alpha)$
(2) $-(\beta+P _{50}(\alpha))$
(3) $P _{50}(\alpha)-\beta$
(4) $\beta+P _{50}(\alpha)$
Show Answer
Answer: (2)
Solution:
Formula: Standard formulas for Indefinite Integration, Properties of definite integral
$I =\int_0^{\alpha} \frac{t^{50}}{1-t} dt= \int_0^{\alpha} \frac{t^{50}-1+1}{1-t}$
$I=-\int_0^{\alpha}(1+t+\ldots . .+t^{49})+\int_0^{\alpha} \frac{1}{1-t} dt$
$I=-(\frac{\alpha^{50}}{50}+\frac{\alpha^{49}}{49}+\ldots . .+\frac{\alpha^{1}}{1})+(\frac{\ln (1-f)}{-1})_0^{\alpha}$
$I=-P _{50}(\alpha)-\ln (1-\alpha)$
$I=-P _{50}(\alpha)-\beta$
