The total area enclosed by the graphs of y=6(x^2)–(x^3)+x and y=(x^2)+7x is ..
graph those two functions to get your limits
The second x(x+7) is a parabola right side up (holds water) with zeros at x = 0 and x = -7
The first x ( -x^2 + 6x + 1) is a cubic with zero at zero.
These two curves intersect at x+7 = -x^2+6x+1
which is at x = 2 and x = 3
Does that help ?
Yeah. Thanks. I found out that the limits are from 0 to 2 and 2 to 3.
To find the total area enclosed by the graphs of the two functions, y=6(x^2)–(x^3)+x and y=(x^2)+7x, we need to find the intersection points of the two graphs and then calculate the area between these intersection points.
Step 1: Find the intersection points:
Set the two equations equal to each other:
6(x^2)–(x^3)+x = (x^2)+7x
Rearrange the equation to get a cubic equation equal to zero:
x^3+5x^2-6x=0
Factor out the common factors:
x(x^2+5x-6)=0
Solve for x by setting each factor equal to zero:
x = 0 or x^2+5x-6 = 0
Solve the quadratic equation x^2+5x-6 = 0 using factoring or the quadratic formula to find the other two values of x.
Step 2: Calculate the area:
Once you have the values of x where the two graphs intersect, you can calculate the area enclosed between the graphs.
Since there are multiple possible intervals of intersection, you need to find the bounds of integration. These bounds are the x-values where the two functions intersect.
Once you have the bounds of integration, you can find the definite integral of the difference between the two functions over this interval to calculate the area enclosed.
The formula for calculating the area between two curves is:
A = ∫[a,b] (f(x) - g(x)) dx
In this case, your a and b values will be the x-values where the two functions intersect, and f(x) and g(x) will be the corresponding y-values of the two functions.
Evaluate the definite integral over the given interval to find the total area enclosed by the graphs of the two functions.
Keep in mind that sometimes the area between two curves may be negative if one curve lies below the other. If this is the case, take the absolute value of the result to get the actual area.
Note: If you're not familiar with calculus and integration, you can also approximate the area using numerical methods such as graphing the functions and counting the squares. However, this method may not give an exact value.