Find the area between the two curves
y=x^4-10x^2+36 and y=3x^2
what's the trouble? You are adding up a bunch of thin rectangles, with width dx and height the distance between the curves. The curves intersect at (±2,12) and (±3,27). Since both functions are even, we can use symmetry and use
a = 2∫[2,3] (3x^2)-(x^4-10x^2+36) dx = 124/15
@Steve, my teacher says the answer is supposed to be 1436/15 but I have to figure out how I got it.
Find the area of the regions that are enclosed by the curves.
surely you can do the math.
∫[2,3] (3x^2)-(x^4-10x^2+36) dx
= ∫[2,3] -x^4 + 13x^2 - 36 dx
= -x^5/5 + 13/3 x^3 - 36x [2,3]
= (-243/5 + 13*27/3 - 36*3)-(-32/5 + 13/3 * 8 - 36*2)
= 62/15
double that for 124/15
Now you need to ask your teacher how 1436/15 can be right. Just looking at the graphs, you can see that the area is very small. 1436/15 is almost 100!
http://www.wolframalpha.com/input/?i=x^4-10x^2%2B36+%3D+3x^2
To find the area between the two curves, we need to find the points where the curves intersect and then integrate the difference between the two curves over the interval of intersection.
Step 1: Find the points of intersection:
We can start by setting the two equations equal to each other:
x^4 - 10x^2 + 36 = 3x^2
Rearranging the equation, we get:
x^4 - 13x^2 + 36 = 0
This is a quadratic equation in terms of x^2. We can solve it by factoring:
(x^2 - 4)(x^2 - 9) = 0
Setting each factor equal to zero, we get:
x^2 - 4 = 0 or x^2 - 9 = 0
Solving each equation, we find two sets of solutions:
x^2 - 4 = 0:
x = ±2
x^2 - 9 = 0:
x = ±3
So the points of intersection are (-3, 9), (-2, 4), (2, 4), and (3, 9).
Step 2: Integrate the difference between the two curves:
To find the area between the two curves, we need to integrate the difference between the curves over the interval of intersection.
Since the curves intersect at x = -3, -2, 2, and 3, we need to integrate over each subinterval.
For the interval x = -3 to x = -2:
∫ (3x^2 - (x^4 - 10x^2 + 36)) dx
Simplifying the expression inside the integral, we get:
∫ (13x^2 - x^4 + 36) dx
Integrating term by term, we get:
(13/3)x^3 - (1/5)x^5 + 36x + C
Evaluating the definite integral from x = -3 to x = -2, we get:
[(13/3)(-2)^3 - (1/5)(-2)^5 + 36(-2)] - [(13/3)(-3)^3 - (1/5)(-3)^5 + 36(-3)]
Repeat the above integration process for each subinterval (x = -2 to x = 2 and x = 2 to x = 3) and add up the results to find the total area between the two curves.