n-IN-I-n-1-e-x-n-1-lnx-dx-1-prove-that-I-n-is-positive-and-increasing-2-using-a-part-by-part-integration-calculate-I-n-

Question Number 143702 by henderson last updated on 17/Jun/21

n \in IN .

I_{n} = \int_{1}^{e} x^{n + 1} l n x d x .

1 . prove that (I_{n}) is positive and increasing .

2 . using a part - by - part integration, calculate I_{n} .

Answered by mindispower last updated on 17/Jun/21

l n (x) ⩾ 0, \forall x ⩾ 1

\Rightarrow \int_{1}^{e} x^{n + 1} l n (x) ⩾ \int_{1}^{e} 0 . d x = 0

2 I_{n} = {[\frac{x^{n + 2}}{n + 2} l n (x)]}_{1}^{e} - \frac{1}{n + 2} \int_{1}^{e} \frac{x^{n + 2}}{x} d x

\frac{e^{n + 2}}{n + 2} - \frac{1}{{(n + 2)}^{2}} {[x^{n + 2}]}_{1}^{e}

= \frac{(n + 1) e^{n + 2} + 1}{{(n + 2)}^{2}}

Answered by mathmax by abdo last updated on 17/Jun/21

1) we have 1 ⩽ x ⩽ e \Rightarrow x^{n + 1} logx ⩾ 0 and

I_{n + 1} - I_{n} = \int_{1}^{e} (x^{n + 2} - x^{n + 1}) logx dx = \int_{1}^{e} x^{n + 1} (x - 1) logx dx ⩾ 0 \Rightarrow

I_{n} is increazing

2) by parts I_{n} = {[\frac{x^{n + 2}}{n + 2} logx]}_{1}^{e} - \int_{1}^{e} \frac{x^{n + 1}}{n + 2} dx

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

= \frac{n + 2 - (e^{n + 2} - 1)}{{(n + 2)}^{2}} = \frac{n + 3 - e^{n + 2}}{{(n + 2)}^{2}}