For n in N,let P_n = int_1^e (ln x)^n dx. The | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $P_n = \int_1^e (\ln x)^n dx$. Using integration by parts,let $u = (\ln x)^n$ and $dv = dx$. Then $du = n(\ln x)^{n-1} \cdot \frac{1}{x} dx$

Vedclass · Accepted Answer

$-9e$

Vedclass · Answer

(B) Given $P_n = \int_1^e (\ln x)^n dx$. Using integration by parts,let $u = (\ln x)^n$ and $dv = dx$. Then $du = n(\ln x)^{n-1} \cdot \frac{1}{x} dx$ and $v = x$. $P_n = [x(\ln x)^n]_1^e - \int_1^e x \cdot n(\ln x)^{n-1} \cdot \frac{1}{x} dx$. $P_n = e(1)^n - 0 - n \int_1^e (\ln x)^{n-1} dx$. $P_n = e - n P_{n-1}$. Now,we need to find $P_{10} - 90P_8$. From the recurrence relation: $P_{10} = e - 10P_9$. $P_9 = e - 9P_8$. Substitute $P_9$ into the expression for $P_{10}$: $P_{10} = e - 10(e - 9P_8)$. $P_{10} = e - 10e + 90P_8$. $P_{10} = -9e + 90P_8$. Therefore,$P_{10} - 90P_8 = -9e$.

For $n \in N$,let $P_n = \int_1^e (\ln x)^n dx$. Then $(P_{10} - 90P_8)$ is equal to

Similar Questions

The value of $I = \int_{\pi / 2}^{5 \pi / 2} \frac{e^{\tan^{-1}(\sin x)}}{e^{\tan^{-1}(\sin x)} + e^{\tan^{-1}(\cos x)}} dx$ is

$\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x=$

$\int_0^\pi x f(\sin x) dx = $

$\int_{\log \frac{1}{2}}^{\log 2} \sin \left(\frac{e^x-1}{e^x+1}\right) d x=$

If $\int \limits_0^1 \frac{1}{\left(5+2 x -2 x ^2\right)\left(1+ e ^{(2-4 x)}\right)} dx =\frac{1}{\alpha} \log _{ e }\left(\frac{\alpha+1}{\beta}\right)$ where $\alpha, \beta > 0$,then $\alpha^4-\beta^4$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module