Let f(x) = intlimits_0^x {(t^2 + 2t + 2)dt} w | vedclass

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(B) Given the inequality: $\log_{\sqrt{2}}(1 + \sqrt{6x - x^2 - 8}) \ge 0$.Since the base $\sqrt{2} > 1$,we have $1 + \sqrt{6x - x^2 - 8} \ge (\sqrt{2

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$56$

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(B) Given the inequality: $\log_{\sqrt{2}}(1 + \sqrt{6x - x^2 - 8}) \ge 0$.Since the base $\sqrt{2} > 1$,we have $1 + \sqrt{6x - x^2 - 8} \ge (\sqrt{2})^0 = 1$.This implies $\sqrt{6x - x^2 - 8} \ge 0$,which is always true for the domain of the square root.For the square root to be defined,$6x - x^2 - 8 \ge 0$,which means $x^2 - 6x + 8 \le 0$.Factoring the quadratic,$(x - 2)(x - 4) \le 0$,so $x \in [2, 4]$.Now,$f(x) = \int_0^x (t^2 + 2t + 2)dt = [\frac{t^3}{3} + t^2 + 2t]_0^x = \frac{x^3}{3} + x^2 + 2x$.Since $f'(x) = x^2 + 2x + 2 = (x+1)^2 + 1 > 0$,$f(x)$ is strictly increasing.Thus,the range of $f(x)$ for $x \in [2, 4]$ is $[f(2), f(4)]$.$a = f(2) = \frac{8}{3} + 4 + 4 = \frac{8 + 24}{3} = \frac{32}{3}$.$b = f(4) = \frac{64}{3} + 16 + 8 = \frac{64 + 72}{3} = \frac{136}{3}$.Therefore,$a + b = \frac{32}{3} + \frac{136}{3} = \frac{168}{3} = 56$.

Let $f(x) = \int\limits_0^x {(t^2 + 2t + 2)dt}$ where $x$ is the set of real numbers satisfying the inequation $\log_{\sqrt{2}}(1 + \sqrt{6x - x^2 - 8}) \ge 0$. If the range of $f(x)$ is $[a, b]$,then $(a + b)$ is:

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