Evaluate-0-3-x-2-3x-x-3-

Question Number 59679 by Forkum Michael Choungong last updated on 13/May/19

E v a l u a t e

\underset{0}{\int^{3}} (\frac{x^{2} + 3 x}{x^{3}})

Answered by Kunal12588 last updated on 13/May/19

\underset{0}{\int^{3}} (\frac{x^{2} + 3 x}{x^{3}}) d x

= \underset{0}{\int^{3}} (\frac{x^{2}}{x^{3}} + \frac{3 x}{x^{3}}) d x

= \int_{0}^{3} \frac{1}{x} d x + 3 \int_{0}^{3} \frac{1}{x^{2}} d x

= {[l n (x)]}_{0}^{3} + 3 {[\frac{x^{- 2 + 1}}{- 2 + 1}]}_{0}^{3}

= l n (3) - l n (0) - 3 (3^{- 1} - 0^{- 1})

s o \int_{1}^{3} (\frac{x^{2} + 3 x}{x^{3}}) d x i s u n d e f i n e d

Answered by meme last updated on 13/May/19

\int_{0}^{3} \frac{x + 3}{x^{2}} = \int_{0}^{3} \frac{1}{x} + \int_{0}^{3} \frac{1}{x^{2}} = {[l n (x)]}_{0}^{3} - 3 {[\frac{1}{x}]}_{0}^{3}

i m p o s s i b l e (l n 0)

Answered by Joel122 last updated on 14/May/19

Let f (x) = \frac{x^{2} + 3 x}{x^{3}}

f (x) is undefined at x = 0, so {it}^{'} s an improper integral

I = \underset{0}{\int^{3}} \frac{x^{2} + 3 x}{x^{3}} d x = \lim_{a \to 0^{-}} [\underset{a}{\int^{3}} \frac{x^{2} + 3 x}{x^{3}} d x]

[\begin{matrix} \underset{a}{\int^{3}} \frac{x^{2} + 3 x}{x^{3}} d x = \underset{a}{\int^{3}} \frac{1}{x} + \frac{3}{x^{2}} d x = {[\ln x - \frac{3}{x}]}_{a}^{3} \\ = \ln (\frac{3}{a}) + \frac{3}{a} - 1 \end{matrix}]

I = \lim_{a \to 0^{-}} [\ln (\frac{3}{a}) + \frac{3}{a} - 1] = \infty

∴ The integral is divergent

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