Find the value of int_{0}^{9} [sqrt{x} + 2] , | vedclass

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

Question

(A) Let $I = \int_{0}^{9} [\sqrt{x} + 2] \,dx$. Since $2$ is an integer,we can write this as $I = \int_{0}^{9} ([\sqrt{x}] + 2) \,dx = \int_{0}^{9} [\

Vedclass · Accepted Answer

$31$

Vedclass · Answer

(A) Let $I = \int_{0}^{9} [\sqrt{x} + 2] \,dx$. Since $2$ is an integer,we can write this as $I = \int_{0}^{9} ([\sqrt{x}] + 2) \,dx = \int_{0}^{9} [\sqrt{x}] \,dx + \int_{0}^{9} 2 \,dx$.First,evaluate $\int_{0}^{9} [\sqrt{x}] \,dx$ by splitting the interval based on the values of $[\sqrt{x}]$:For $0 \le x For $1 \le x For $4 \le x Thus,$\int_{0}^{9} [\sqrt{x}] \,dx = \int_{0}^{1} 0 \,dx + \int_{1}^{4} 1 \,dx + \int_{4}^{9} 2 \,dx = 0 + (4 - 1) + 2(9 - 4) = 3 + 10 = 13$.Next,$\int_{0}^{9} 2 \,dx = [2x]_{0}^{9} = 18$.Therefore,$I = 13 + 18 = 31$.

Find the value of $\int_{0}^{9} [\sqrt{x} + 2] \,dx$,where $[.]$ denotes the greatest integer function.

Similar Questions

By the definition of the definite integral,the value of $\lim _{n \rightarrow \infty}\left(\frac{1^4}{1^5+n^5}+\frac{2^4}{2^5+n^5}+\frac{3^4}{3^5+n^5}+\ldots+\frac{n^4}{n^5+n^5}\right)$ is

The value of $\int_{-1}^{1} \frac{dx}{\sqrt{|x|}}$ is

$\int_0^{2/3} \frac{dx}{4 + 9x^2} = $

$\int \limits_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^2}} d x$ is equal to

$f(x) = \begin{cases} x^2, & 0 \leq x < 1 \\ \sqrt{x}, & 1 \leq x \leq 2 \end{cases} \implies \int_0^2 f(x) \, dx = ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module