Find the area of region between the graph of y=x^2+4x+3 and y=x^2 between the intervals x=1 and x=3

Answer 1

To find the area of the region between the two graphs, we need to find the definite integral of the absolute difference between the two functions within the given intervals.

First, let's find the points of intersection between the two curves.
Setting the two equations equal to each other, we have:
x^2 + 4x + 3 = x^2.
This simplifies to:
4x + 3 = 0,
4x = -3,
x = -3/4.

The area between the two curves can be expressed as:
∫[1, 3] |(x^2 + 4x + 3) - x^2| dx.

Simplifying the absolute difference, we get:
∫[1, 3] |4x + 3| dx.

Now, let's consider the two cases for the inner absolute value function, 4x + 3:
1. If 4x + 3 ≥ 0, the integrand is 4x + 3.
2. If 4x + 3 < 0, the integrand is -(4x + 3) = -4x - 3.

Thus, our integral becomes:
∫[1, 3] (4x + 3) dx + ∫[1, 3] (-4x - 3) dx.

Integrating each term:
∫[1, 3] (4x + 3) dx = [2x^2 + 3x] from 1 to 3,
= [2(3)^2 + 3(3)] - [2(1)^2 + 3(1)],
= [18 + 9] - [2 + 3],
= 27 - 5,
= 22.

And,
∫[1, 3] (-4x - 3) dx = [-2x^2 - 3x] from 1 to 3,
= [-2(3)^2 - 3(3)] - [-2(1)^2 - 3(1)],
= [-18 - 9] - [-2 - 3],
= -27 + 5,
= -22.

The area between the two graphs is given by the sum of the absolute values of the integrals:
Area = |22| + |-22| = 22 + 22 = 44.

Therefore, the area of the region between the two graphs is 44 square units.

Answer 2

the curves do not intersect on the given interval. So the area between them is just the simple
∫[1, 3] (4x + 3) dx = 22

Answer 3

Apologies for the incorrect information in my previous response. You are correct, the curves do not intersect on the given interval. Therefore, the area between them is simply given by the integral of (4x + 3) from x = 1 to x = 3.

∫[1, 3] (4x + 3) dx = [2x^2 + 3x] from 1 to 3
= (2(3)^2 + 3(3)) - (2(1)^2 + 3(1))
= (18 + 9) - (2 + 3)
= 27 - 5
= 22

The area between the two graphs is 22 square units.