Question

ꇙꄲ꒒꒦ꏂ ꒐꓄....꒯ꄲꋊ'꓄ ꇙꉣꋬꂵ....​

Answer 1

Let us define a function such that,

$\longrightarrow g(x)=\dfrac{x}{120}+\dfrac{x^3}{30}$

Its derivative wrt $x$ will be,

$\longrightarrow g'(x)=\dfrac{1}{120}+\dfrac{x^2}{10}$

Here $g'(x)>0$ for every $x.$

Hence $g(x)$ is strictly increasing.

Our function is,

$\longrightarrow f(x)=\left[\dfrac{x}{120}+\dfrac{x^3}{30}\right]$

$\longrightarrow f(x)=[g(x)]$

For $x=0,$

$\longrightarrow f(0)=\left[\dfrac{0}{120}+\dfrac{0^3}{30}\right]$

$\longrightarrow f(0)=0$

For $x=3,$

$\longrightarrow f(3)=\left[\dfrac{3}{120}+\dfrac{3^3}{30}\right]$

$\longrightarrow f(3)=\left[\dfrac{37}{40}\right]$

$\longrightarrow f(3)=0$

Therefore, since $g$ is strictly increasing,

$\longrightarrow f(x)=0\quad\!\forall x\in[0,\ 3]$

Then,

$\displaystyle\longrightarrow\int\limits_{f(x)}^{f(x)+1}(f(x)+2)\ dx=\int\limits_{0}^{0+1}(0+2)\ dx$

$\displaystyle\longrightarrow\int\limits_{f(x)}^{f(x)+1}(f(x)+2)\ dx=\int\limits_0^12\ dx$

$\displaystyle\longrightarrow\underline{\underline{\int\limits_{f(x)}^{f(x)+1}(f(x)+2)\ dx=2}}$

Hence 2 is the answer.