Animated Solution for Mathematics - Definite Integration: Let the line $x=-1$ divide the area of the region $\{(x,y):1+x^2\le y\le 3-x\}$ in the ratio $m:n,\text{gcd}(m,n)=1$. Then $m+n$ is equal to

Imagine you are standing on a coordinate plane, looking at two distinct mathematical entities. On one hand, you have the curve $y=1+x^2$, a classic, upward-opening parabola that has been gracefully lifted by one unit.

On the other, you have the line $y=3-x$, a straight, downward-sloping path. These two shapes define a region trapped between them. A vertical line, $x=-1$, cuts through this region, dividing it into two parts. Our mission is to find the ratio of these two areas.

Before we can calculate anything, we must determine where this region begins and ends by finding the intersection points. At these points, the 'ceiling' (the line) and the 'floor' (the parabola) meet.

We set them equal:

Rearranging this, we get the quadratic equation:

Factoring this yields $(x+2)(x-1)=0$. This tells us that our region starts at $x=-2$ and ends at $x=1$. These are the absolute limits of our integration.

The line $x=-1$ enters the scene and slices our total region into two parts: $A_1$ on the left and $A_2$ on the right. $A_1$ is the area from $x=-2$ to $x=-1$, and $A_2$ is the area from $x=-1$ to $x=1$.

To find the area of any slice, we integrate the difference between the upper curve and the lower curve:

In our case, the line is always on top, so the integrand is $(3-x)-(1+x^2)$, which simplifies to:

Let's tackle $A_1$ first. We integrate $2-x-x^2$ from $-2$ to $-1$. The antiderivative is:

Evaluating this at the limits:

Now, for $A_2$, we integrate the same expression from $-1$ to $1$:

We have our two areas: $A_1=\frac{7}{6}$ and $A_2=\frac{10}{3}$. To find the ratio $m:n$, we compare them.

Converting $A_2$ to have a denominator of $6$, we get $\frac{20}{6}$. Thus, the ratio is:

Since the greatest common divisor of $7$ and $20$ is $1$, we have $m=7$ and $n=20$. The final step is to find $m+n$: