Question

The mean value of the function ƒ(x) = $\frac{1}{x^{2} + x}$on the interval [1, 3/2] is –

• A. log (6/5)
• B. 2 log (6/5)
• C. 4
• D. log 3/5

Answer 1

Answer: (B)

2 log (6/5)

Answer 2

Mean value = $\frac { 1 } { \mathrm {~b} - \mathrm { a } }$ $\int_{a}^{b}{ƒ(x)dx}$

(Property 14)

= $\frac { 1 } { 3 / 2 - 1 }$ $\int_{1}^{3/2}\frac{1}{x^{2} + x}$dx = 2 $\int_{1}^{3/2}\left\lbrack \frac{1}{x} - \frac{1}{x + 1} \right\rbrack$dx

= 2 (log x – log (x $\left. \ + 1) \right|_{1}^{3/2}$= 2 [log (3/2) – log

(5/2) – (log 1 – log 2)]

= 2 log (6/5)