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 the integral $\int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} d x$ is :

The value of $\frac{8}{\pi} \int \limits_0^{\frac{\pi}{2}} \frac{(\cos x)^{2023}}{(\sin x)^{2023}+(\cos x)^{2023}} dx$ is $.............$.

The value of $\int_{-2}^{2}\left|3 x^{2}-3 x-6\right| d x$ is ...... .

The value of the integral $I = \int_{0}^{1} x(1 - x)^n dx$ is

Evaluate the definite integral: $\int_{\frac{1}{2}}^{2} \frac{x^2 \ln x}{(1+x^2)^3} dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module