area !!!!!!

$\begin{aligned} I & = \int_1^{37} g(x) \ dx & \small \color{#3D99F6} \text{Let }y = f(x) = x^3+3x+1 \\ & = \int_1^{37} x \ dy & \small \color{#3D99F6} \text{Note: }\begin{cases} y = 1 & \implies x = 0 \\ y =37 & \implies x=3 \end{cases} \\ & = \int_0^3 x \color{#3D99F6} \left(3x^2+3\right) dx & \small \color{#3D99F6} \text{and }dy = (3x^2 + 3) \ dx \\ & = \int_0^3 3\left(x^3+x\right) dx \\ & = 3 \left[\frac {x^4}4+\frac {x^2}2 \right]_0^3 \\ & = \boxed{74.25} \end{aligned}$

Since $g\left( x \right)$ is inverse of $f\left( x \right)$

Therfore $f\left( x \right)$ is a mirror image of $g\left( x \right)$ to mirror $y=x$

Let initially a graph contains $x=1,\quad x=37$ lines and $y=g(x)$ curve

After take the mirror images of all lines and curves about $y=x$

The graph now contains $y=1,\quad y=37$ lines and $y=f(x)$ curve as shown

(Note:Since f'(x) is positive hence graph is always increasing and hence gives one x-coordinate for one y-coordinate)

By solving $f(x)=1$ & $f(x)=37$ seperately we get

$x=0$ for $f(x)=1$ , and, $x=3$ for $f(x)=37$

Let O(0,1), A(0,37) and B(3,37) be three points

Here area of graph $f(x)$ between $y=1$ & $y=37$ with y-axis is our answer which is equal to:

$area(OAB)=area(rectangle\quad by\quad x=0,y=0,x=3\quad \& \quad y=37)+area(f(x)\quad with\quad x-axis\quad from\quad x=0\quad to\quad x=3)$ - eq(i)

Therefore area of f(x) with x-axis from x=0 to x=3 is

$\int _{ 0 }^{ 3 }{ f(x)dx }$

$\frac { { 3 }^{ 4 } }{ 4 } +\frac { 3{ (3) }^{ 2 } }{ 2 } +3$

And area of rectangle ans in equation (i) is

$37\quad X\quad 3$

And then using equation (i) we can get our answer

$\ast$ $g(x)$ is the inverse of $f(x)$ .

$\ast$ $f'(x) = 3x^2+3 >0$ for all $x$ , so $f(x)$ and $g(x)$ are strictly increasing.(Symmetric about the line $y=x$ ).

$\ast f(0) =1$ then $g(1) = 0$

Therefore, $g(x) \geq 0$ for $x \in [1,37]$ Therefore our area is:

$\displaystyle A = \int_1^{37} g(x) dx$

Let $u= g(x)$ :

$\displaystyle x=f(u) => dx = f'(u) du$

And since $f(0) =1 , f(3) = 37$ , the integral becomes:

$\displaystyle A= \int_0^3 uf'(u) du$

$\displaystyle = \int_0^3 u(3u^2+3u) du$

The rest is easy work that will lead to the final answer:

$\displaystyle \boxed{A= 74.25 }$