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
Find the area between the two curves
y=x^4-10x^2+36 and y=3x^2
3 answers
@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.
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
∫[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
1 answer