Evaluate the definite integral int_{0}^{9} [s | vedclass

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(A) We are given the integral $I = \int_{0}^{9} [\sqrt{x} + 2] \, dx$. Using the property of the Greatest Integer Function $[x + n] = [x] + n$ for any

Vedclass · Accepted Answer

$31$

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(A) We are given the integral $I = \int_{0}^{9} [\sqrt{x} + 2] \, dx$. Using the property of the Greatest Integer Function $[x + n] = [x] + n$ for any integer $n$,we have $[\sqrt{x} + 2] = [\sqrt{x}] + 2$. Thus,$I = \int_{0}^{9} [\sqrt{x}] \, dx + \int_{0}^{9} 2 \, dx$. Evaluating the second part: $\int_{0}^{9} 2 \, dx = 2[x]_{0}^{9} = 2(9 - 0) = 18$. For the first part,we split the interval $[0, 9]$ based on the values where $\sqrt{x}$ is an integer: $\int_{0}^{9} [\sqrt{x}] \, dx = \int_{0}^{1} [\sqrt{x}] \, dx + \int_{1}^{4} [\sqrt{x}] \, dx + \int_{4}^{9} [\sqrt{x}] \, dx$. In $[0, 1)$,$[\sqrt{x}] = 0$. In $[1, 4)$,$[\sqrt{x}] = 1$. In $[4, 9)$,$[\sqrt{x}] = 2$. So,$\int_{0}^{9} [\sqrt{x}] \, dx = \int_{0}^{1} 0 \, dx + \int_{1}^{4} 1 \, dx + \int_{4}^{9} 2 \, dx$. $= 0 + (4 - 1) + 2(9 - 4) = 3 + 2(5) = 3 + 10 = 13$. Finally,$I = 13 + 18 = 31$.

Evaluate the definite integral $\int_{0}^{9} [\sqrt{x} + 2] \, dx$,where $[\cdot]$ denotes the Greatest Integer Function $(G.I.F.)$.

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