If $\int_{\mathrm{n}}^{\mathrm{n}+1} \mathrm{~g}(\mathrm{x}) \mathrm{dx}=\mathrm{n}^2, \forall \mathrm{n}…
If $\int_{\mathrm{n}}^{\mathrm{n}+1} \mathrm{~g}(\mathrm{x}) \mathrm{dx}=\mathrm{n}^2, \forall \mathrm{n} \in \mathbb{Z}$, then the value of $\int_{-3}^3 \mathrm{~g}(\mathrm{x}) \mathrm{dx}$ is
$19$
$28$
$9$
$27$
Solution
Given, $\int_n^{n+1} g(x) d x=n^2, \forall n \in Z$
at $n=-3, \int_{-3}^{-2} g(x) d x=9$ ...(i)
at $\mathrm{n}=-2, \int_{-2}^{-1} g(x) d x=4$ ...(ii)
at $\mathrm{n}=-1, \int_{-1}^0 g(x) d x=1$ ...(iii)
at $\mathrm{n}=0, \int_0^1 g(x) d x=0$ ...(iv)
at $\mathrm{n}=1 \int_1^2 g(x) d x=1$ ...(v)
at $\mathrm{n}=2 \int_2^3 g(x) d x=4$ ...(vi)
Adding eqs. (i), (ii), (iii), (iv), (v) \& (vi), we get
$
\begin{aligned}
& \Rightarrow \int_{-3}^{-2} g(x) d x+\int_{-2}^{-1} g(x) d x+\int_{-1}^0 g(x) d x+\int_0^1 g(x) d x+\int_1^2 g(x) d x \\
& +\int_2^3 g(x) d x=19 \\
& \Rightarrow \int_{-3}^3 g(x) d x=19 \\
&
\end{aligned}
$