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

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(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.

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