Find the area enclosed by y=x^2-2x-3 and y=x+1

I got 17/3 is this right? If not how is this problem solved?

Here is the problem solved on a piece of paper! sorry for being lat but i was looking for my camera :D

http://imageshack.us/f/684/imgfffffff978.jpg/

ENJOY!

A picture really helps. You'll be integrating in y from the lower function (the parabola) to the upper function (the line).

You'll be integrating in x from one point of intersection of the parabola and the line to the other.

Solve the system to find those points.

y = x^2 - 2x - 3

y = x + 1

x^2 - 2x - 3 = x + 1

Simplify.

x^2 - 3x - 4 = 0

(x - 4)(x + 1) = 0

x = 4, x = - 1


Set up your double definite integral:

A = integral [- 1, 4] integral [x + 1, x^2 - 2x - 3] {dydx}

Integrating dy first

A = integral [- 1, 4] {y} [x + 1, x^2 - 2x - 3] dx

Evaluating the limits of y

A = integral [- 1, 4] {x + 1 - (x^2 - 2x - 3)} dx

Simplify

A = integral [- 1, 4] {x + 1 - x^2 + 2x + 3} dx

Simplify

A = integral [- 1, 4] {- x^2 + 3x + 4} dx

Evaluate the second integral

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

Apply the limits.

A = {(- 1/3)4^3 + (3/2)4^2 + 4(4)} - {(- 1/3)(- 1)^3 + (3/2)(- 1)^2 + 4(- 1)}

Simplify.

A = {- 64/3 + 48/2 + 16} - { (1/3) + (3/2) - 4}

Simplify

A = {- 64/3 + 40} - {- 13/6}

Simplify

A = 125/6

First find the intersection pts:

x² - 2x - 3 = x + 1

x² - 3x - 4 = 0

(x - 4)(x + 1) = 0

x = 4, -1

Area = ∫(x + 1 - x² + 2x + 3) dx from -1 to 4

= ∫(-x² + 3x + 4) dx from -1 to 4

= -x^3/3 + (3x^2)/2 + 4x eval. from -1 to 4

= -64/3 + 24 + 16 - 1/3 - 3/2 + 4 = -65/3 + 44 - 3/2 = 125/6 sq. units

da=[(x+1)-(x²-2x-3)]dx
=[-x²+3x+4]dx
4
area=∫(4+3x-x²)dx=[4x+3x²/2-x³/3] from -1 to 4
-1 = [16+24-64/3]-[-4+3/2+1/3]
=44-(64/3+3/2+1/3)
=20 5/6
=20.833