If int_{-a}^{a} (|x| + |x-2|) dx = 22,(a 2) | vedclass

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(B) Given $\int_{-a}^{a} (|x| + |x-2|) dx = 22$ with $a > 2$.We split the integral based on the critical points $x=0$ and $x=2$:$\int_{-a}^{0} (-x - (

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$3$

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(B) Given $\int_{-a}^{a} (|x| + |x-2|) dx = 22$ with $a > 2$.We split the integral based on the critical points $x=0$ and $x=2$:$\int_{-a}^{0} (-x - (x-2)) dx + \int_{0}^{2} (x - (x-2)) dx + \int_{2}^{a} (x + x-2) dx = 22$$\int_{-a}^{0} (-2x + 2) dx + \int_{0}^{2} 2 dx + \int_{2}^{a} (2x - 2) dx = 22$$[-x^2 + 2x]_{-a}^{0} + [2x]_{0}^{2} + [x^2 - 2x]_{2}^{a} = 22$$(0 - (-a^2 - 2a)) + (4 - 0) + ((a^2 - 2a) - (4 - 4)) = 22$$a^2 + 2a + 4 + a^2 - 2a = 22$$2a^2 + 4 = 22 \Rightarrow 2a^2 = 18 \Rightarrow a^2 = 9$. Since $a > 2$,$a = 3$.Now,we evaluate $\int_{3}^{-3} (x + [x]) dx = -\int_{-3}^{3} (x + [x]) dx$.$\int_{-3}^{3} x dx + \int_{-3}^{3} [x] dx = 0 + ([-3] + [-2] + [-1] + [0] + [1] + [2])$$= -3 - 2 - 1 + 0 + 1 + 2 = -3$.Thus,$-\int_{-3}^{3} (x + [x]) dx = -(-3) = 3$.

If $\int_{-a}^{a} (|x| + |x-2|) dx = 22$,$(a > 2)$ and $[x]$ denotes the greatest integer $\leq x$,then $\int_{a}^{-a} (x + [x]) dx$ is equal to ...........

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