Question

Let $\alpha >3$ be a fixed number. Evaluate the improper integral ${\int }_0^{\alpha }\frac{1}{(x-3{)}^2}dx$

• A. $\frac{1}{\alpha -3}$
• B. $0$
• C. $\infty$
• D. $\frac{1}{(\alpha +3{)}^2}$
• E. $\frac{1}{(\alpha -3{)}^2}$

Answer 1

Answer: (A)

$\frac{1}{\alpha -3}$

Given : ${\int }_0^{\alpha }\frac{1}{(x-3{)}^2}$

$I={\int }_0^{\alpha }\frac{dx}{(x-3{)}^2}={[-\frac{1}{(x-3)}]}_0^{\alpha }=-\frac{1}{(\alpha -3)}+\frac{1}{-3}=-\frac{1}{3}-\frac{1}{(\alpha -3)}$

Hence the correct answer is $-\frac{1}{3}-\frac{1}{(\alpha -3)}$