How to Find the Area Between Two Curves for Functions of y: Step-by-Step Guide with Example Problem

area between 2 curves (for functions of y)

(right curve – left curve)

To find the area between two curves for functions of y, follow these steps:

1. Sketch the graphs of the given functions to identify their points of intersection. Let’s assume that we have two functions, f(y) and g(y), and we want to find the area between them. We can start by graphing them and identifying their points of intersection.
2. Determine the limits of integration. Once we have identified the points of intersection, these will be the limits of our integral. For example, if the two functions intersect at two points (y1, x1) and (y2, x2), then our limits of integration would be from y1 to y2.
3. Determine which function is on top. We need to determine which function is on top at each value of y in the interval that we are integrating over. We can do this by comparing the two functions at each value of y. The function that is higher up (i.e., has a greater y-value) is the one that is on top.
4. Set up the integral. To find the area between the two curves, we need to integrate the difference between the two functions with respect to y. The integral will be the integral of f(y) – g(y) from y1 to y2.
5. Evaluate the integral. Once we have set up the integral, we can evaluate it using integration techniques such as substitution or integration by parts.

Here is an example problem:

Find the area between the curves y = x^2 + 1 and y = 3x – 2.

1. Sketch the graphs of the two functions:


We can see that the two functions intersect at (1, 2) and (-1/2, 5/4).

2. Determine the limits of integration:

The limits of integration will be from -1/2 to 1.

3. Determine which function is on top:

At y values between 5/4 and 2, the function y = x^2 + 1 is on top. At y values between 2 and 3, the function y = 3x – 2 is on top.

4. Set up the integral:

The integral will be the integral of f(y) – g(y) from 5/4 to 2, plus the integral of g(y) – f(y) from 2 to 3:

∫(x^2 + 1 – (3x – 2)) dy from 5/4 to 2 + ∫((3x – 2) – (x^2 + 1)) dy from 2 to 3

5. Evaluate the integral:

Evaluating the integral using integration by substitution gives us an area of 17/6 square units.

Therefore, the area between the curves y = x^2 + 1 and y = 3x – 2 is 17/6 square units.

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