The region bounded by the graphs of y=9-x^2, y=5-3x find the area?

Answer 1

In order to find the area, we must determine
(1) where they intersect to obtain our bounds for integration, and
(2) which is the higher curve.

Our formula is going to be

A = Integral (a to b, f(x) - g(x) ) dx

where f(x) is our "higher curve".

First, let's obtain the points of intersection. To do this, we have to equate them to each other.

9 - x^2 = 5 - 3x

Moving everything over to the right hand side, we get

0 = x^2 + 5 - 3x - 9
0 = x^2 - 3x - 4
0 = (x - 4) (x + 1)

Therefore, x = {-1, 4}

Our bounds of integration are -1 and 4.

To determine which curve is the higher curve, test a point in between -1 and 4; use BOTH functions, and whichever yields the higher value is going to be our f(x).

Test x = 0: Then, for the first function, y = 9 - 0^2 = 9, and
for the second function, y = 5 - 3(0) = 5. Therefore, the first function is clearly higher, and our area is

A = Integral (-1 to 4, [9 - x^2] - [5 - 3x])dx

A = Integral (-1 to 4, [4 - x^2 + 3x]) dx

Now, we can easily integrate this using the reverse power rule.

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

Plugging in 4, and subtracting it with plugging in -1, we get

A = [4(4) - (4^3)/3 + 3(4^2 / 2)] - [4(-1) - ([-1]^3)/3 + 3( [-1]^2 / 2)]

Now we simplify this.

A = [16 - 64/3 + 3(16/2)] - [-4 - (-1/3) + 3(1/2)]
A = [16 - 64/3 + 48/2] - [-4 + (1/3) + 3/2]
A = [16 - 64/3 + 48/2] - [-4 + (1/3) + 3/2]

See how the first term in each brackets are whole numbers, the second term in each brackets has 3 as a denominator, and the third term has 2? Instead of finding a common denominator for everything, how about we just join them componentwise?

A = [16 - (-4)] + [-64/3 - 1/3] + [48/2 - 3/2]
A = [20] + [-65/3] + [45/2]

Now, let's put this under a common denominator.

A = 120/6 - 130/6 + 135/6

A = -10/6 + 135/6 = 125/6

Answer 2

The region bounded by the graphs of y=9-x^2, y=5-3x find the area?
Thes two graphs intersect when 9-x^2 =5-3x
0 = x^2-3x-4
(x-4)(x+1)= 0
x=4 and x= -1
integral 9-x^2 = 9x -x^3/3 from -1 to 4
= 9(4) -64/3 -(-9 -(-1/3) = 24
integral 5-3x = 5x -3/2x^2 = from -1 to 4
= 20 -1.5*16 -(-5 -1.5 )= 2.5
So area = 24-2.5=21.5