Find the area on the first quadrant (QI) and  under the curve given by 𝑥=𝑦2+1from its intersection with the x-axis to 𝑥=10.

To solve this problem, we need to use the concept of integration in calculus. Here are the steps:

1. First, we need to find the intersection of the curve 𝑥=𝑦^2+1 with the x-axis. The x-axis is defined by 𝑦=0. So, we substitute 𝑦=0 into the equation of the curve to get 𝑥=0^2+1=1. So, the curve intersects the x-axis at 𝑥=1.

2. Next, we need to find the area under the curve from 𝑥=1 to 𝑥=10. This is done by integrating the function that defines the curve over this interval. However, our function is in terms of y, so we need to change it to be in terms of x. We can do this by solving the equation 𝑥=𝑦^2+1 for y to get y = sqrt(x-1).

3. Now, we can integrate the function y = sqrt(x-1) from 𝑥=1 to 𝑥=10. The integral of sqrt(x-1) is (2/3)(x-1)^(3/2). Evaluating this from 𝑥=1 to 𝑥=10 gives us (2/3)(10-1)^(3/2) - (2/3)*(1-1)^(3/2) = (2/3)*9^(3/2) = 18sqrt(3).

So, the area under the curve 𝑥=𝑦^2+1 from its intersection with the x-axis to 𝑥=10 in the first quadrant is 18sqrt(3) square units.