if the curves y = 2x^2+5 and y = 3x^2+1 intersects at the points (2,13) find the area trapped between the?

Question

the two curves and the y axis

Answer 1

The graphs actually intercest at two points, (2, 13) and (-2,13)

Take your limits of integration then as [-2,2]

Graphing the two functions, 3x^2 + 1 is below 2x^2 + 5, so subrtact the first from the second:

Integral { 2x^2 + 5 - 3x^2 - 1} on [-2,2]
=Integral {4 - x^2} on [-2,2]
= {4x - (1/3)x^3} on [-2,2]

={4(2) - (1/3)(2)^3} - {4(-2) - (1/3)(-2)^3}
={8 - (8/3)} - {- 8 + (8/3) }
= 16 - (16/3)
= 32/3

Answer 2

Integrate (3x^2+1 - 2x^2 -5) from -2 to 2

That's x^2-4

Its integral is x^3/3 - 4x

So, area is |2(8/3 - 8)| = |2*-16/3| = 32/3

Answer 3

K = 2ch/3

At the point where the intercept, you get a horizontal line at 13, so that is the cut off point

y = 2x^2 + 5
h = 13
c = 4

K = (2 * 4 * 12)/3
K = (104/3)
K = about 34.67

y = 3x^2 + 1
c = 4
h = 8

K = (2 * 4 * 7)/3
K = (56/3)
K = 18.67

34.67 - 18.67 = 16

ANS : 16

Answer 4

Make definite integration between x-values 0 and 2 for those two functions, subtract the lower one from the upper one and YOu will get the requested area.

If You need instructions how to do it, e-mail me and I will send the complete procedure for solving this problem.

Answer 5

y = 2x^2 + 5; integrate
y(int) = (2/3)x^3 + 5x
A = (2/3)(2)^3 + 5(2) = 8/3 + 10 = 38/3

y = 3x^2 + 1
y(int) = x^3 + x
A = (2)^3 + 2 = 10

A1 - A2 = 8/3

The trapped area is 8/3.

I think - it's been awhile.

Answer 6

area=mod. integral frm2to13of y1-y2