Let [t] denote the greatest integer function. | vedclass

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(A) We evaluate the integral $\int_0^{2.4} [x^2] dx$ by splitting the interval based on the values of $[x^2]$.$\int_0^{2.4} [x^2] dx = \int_0^1 [x^2]

Vedclass · Accepted Answer

$6$

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(A) We evaluate the integral $\int_0^{2.4} [x^2] dx$ by splitting the interval based on the values of $[x^2]$.$\int_0^{2.4} [x^2] dx = \int_0^1 [x^2] dx + \int_1^{\sqrt{2}} [x^2] dx + \int_{\sqrt{2}}^{\sqrt{3}} [x^2] dx + \int_{\sqrt{3}}^2 [x^2] dx + \int_2^{\sqrt{5}} [x^2] dx + \int_{\sqrt{5}}^{2.4} [x^2] dx$$= \int_0^1 0 dx + \int_1^{\sqrt{2}} 1 dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 dx + \int_{\sqrt{3}}^2 3 dx + \int_2^{\sqrt{5}} 4 dx + \int_{\sqrt{5}}^{2.4} 5 dx$$= 0 + (\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) + 3(2 - \sqrt{3}) + 4(\sqrt{5} - 2) + 5(2.4 - \sqrt{5})$$= \sqrt{2} - 1 + 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3} + 4\sqrt{5} - 8 + 12 - 5\sqrt{5}$$= ( -1 + 6 - 8 + 12 ) + (1 - 2)\sqrt{2} + (2 - 3)\sqrt{3} + (4 - 5)\sqrt{5}$$= 9 - \sqrt{2} - \sqrt{3} - \sqrt{5}$Comparing this with $\alpha + \beta \sqrt{2} + \gamma \sqrt{3} + \delta \sqrt{5}$,we get $\alpha = 9$,$\beta = -1$,$\gamma = -1$,and $\delta = -1$.Therefore,$\alpha + \beta + \gamma + \delta = 9 - 1 - 1 - 1 = 6$.

Let $[t]$ denote the greatest integer function. If $\int_0^{2.4} [x^2] dx = \alpha + \beta \sqrt{2} + \gamma \sqrt{3} + \delta \sqrt{5}$,then $\alpha + \beta + \gamma + \delta$ is equal to $..............$.

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