If a in Z^{+},[x] is the greatest integer not | vedclass

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(B) Given the integral $\int_0^a 2^{[x]} dx = 127$,where $a \in Z^{+}$.We can split the integral into unit intervals:$\int_0^1 2^{[x]} dx + \int_1^2 2

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$7$

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(B) Given the integral $\int_0^a 2^{[x]} dx = 127$,where $a \in Z^{+}$.We can split the integral into unit intervals:$\int_0^1 2^{[x]} dx + \int_1^2 2^{[x]} dx + \dots + \int_{a-1}^a 2^{[x]} dx = 127$Since $[x] = k$ for $x \in [k, k+1)$,the integral becomes:$\int_0^1 2^0 dx + \int_1^2 2^1 dx + \int_2^3 2^2 dx + \dots + \int_{a-1}^a 2^{a-1} dx = 127$Evaluating each term:$2^0(1-0) + 2^1(2-1) + 2^2(3-2) + \dots + 2^{a-1}(a-(a-1)) = 127$$2^0 + 2^1 + 2^2 + \dots + 2^{a-1} = 127$This is a geometric progression with $n = a$ terms,first term $1$,and common ratio $2$:$\frac{1(2^a - 1)}{2 - 1} = 127$$2^a - 1 = 127$$2^a = 128$$2^a = 2^7$Therefore,$a = 7$.

If $a \in Z^{+}$,$[x]$ is the greatest integer not more than $x$,and $\int_0^a 2^{[x]} dx = 127$,then $a =$

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