91Ó°ÊÓ

When we talk about integration with respect to x, we're focusing on finding the area between two curves by blending along the x-axis. In simple terms, we calculate the areas at each slice, or vertical strip, parallel to the y-axis.
To find such an area for our exercise between the curves given by the equations \( y = x^2 \) and \( y = 6 - x \), we first determine the intersection points, as shown in Step 1. These points are \( x = -3 \) and \( x = 2 \). These represent the limits of our integration.
The actual calculation involves setting up a definite integral to sum up these areas between the intersection points. We achieve this by taking the difference of the top function \( (6-x) \) and the bottom function \( (x^2) \), thus: \[ \int_{{-3}}^{{2}} (6-x-x^2) \, dx \] Evaluating this integral will tell us the total area between the two curves. By visualizing it, think of each small slice being added up from the left intersection point to the right intersection point.

Integrating with respect to y takes a horizontal slice of the area between the curves. This method might seem more complex at first because it involves rearranging the original functions, but it can be advantageous in certain cases.
For the given functions \( y = x^2 \) and \( y = 6 - x \), we rearrange them to express x explicitly in terms of y. We find \( x = \sqrt{y} \) and \( x = 6 - y \). These expressions allow us to integrate horizontally.
Now, we must flip our perspective. We're now seeing each slice run along fixed y-values. The limits for this integration are from the smaller y-value of the intersection points, 4, to the larger y-value, 9. Our setup for the definite integral becomes: \[ \int_{{4}}^{{9}} (\sqrt{y} - (6-y)) \, dy \] Through this approach, we're essentially looking at the width of each slice generated by the difference \( \sqrt{y} - (6-y) \), from top to bottom across the y-axis limits.

Intersection points can be thought of as the spots where functions "meet" or cross over each other on a graph. They play a crucial role in defining the limits for our integration processes.
To find these intersection points for the curves \( y = x^2 \) and \( y = 6 - x \), we set the equations equal to each other: \[ x^2 = 6 - x \] Solving this quadratic equation, we find \( x = 2 \) and \( x = -3 \). To fully define the points of intersection, we then substitute these x-values back into either of the original equations to find the corresponding y-values. This results in the intersection points \((-3, 9)\) and \((2, 4)\).
These points are critical because they determine the start and end of the "slices" we sum up to find the area between the curves. Without them, we wouldn't know the bounds for our definite integral, whether we're integrating with respect to x or y.