Find the value of ∫09 [x+2]dx where [.] is the greatest integer function

A
31
B
23
C
22
D
27

Solution:

$\begin{array}{l} \int_{0}^{9} [\sqrt{x} + 2] dx \\ = \int_{0}^{1} [\sqrt{x} + 2] dx + \int_{1}^{4} [\sqrt{x} + 2] dx + \int_{4}^{9} [\sqrt{x} + 2] dx \\ = \int_{0}^{1} 2 dx + \int_{1}^{4} (1 + 2) dx + \int_{4}^{9} (2 + 2) dx \\ = 2 + 9 + 20 = 31 \end{array}$

Related content

JEE Advanced 2023

NEET Rank Assurance Program | NEET Crash Course 2023

JEE Main 2023 Question Papers with Solutions

JEE Main 2024 Syllabus

Best Books for JEE Main 2024

JEE Advanced 2024: Exam date, Syllabus, Eligibility Criteria

JEE Main 2024: Exam dates, Syllabus, Eligibility Criteria

JEE 2024: Exam Date, Syllabus, Eligibility Criteria

NCERT Solutions For Class 6 Maths Data Handling Exercise 9.3

JEE Crash Course – JEE Crash Course 2023