Find the area between the curves, f(x)=4-x^2 and g(x)=3x?

Answer 1

First we must find where the curves meet.
Solve 4 - x^2 = 3x
x^2 + 3x - 4 = 0
(x+4)(x-1) = 0
x = -4, 0

So we must integrate f(x) - g(x) from -4 to 0.
Actually it should be absolute value of f(x) - g(x), but a quick sketch shows that f > g in this interval, so f-g is positive.

f - g = 4 - 3x - x^2
Integral value = [4x - 3(x^2)/2 - (x^3)/3] from -4 to 0
= 0 - [-16 - 24 + 64/3]
= 56/3

Answer 2

Let's find two points where the curves intersect:

4-x^2 = 3x // + x^2 - 4

x^2 + 3x - 4 = 0

x^2 + 2*1.5x + 1.5^2 -4 - 1.5^2 = 0

(x + 1.5)^2 - 4 - 2.25 = 0

(x + 1.5)^2 - 6.25 = 0

(x + 1.5)^2 - 2.5^2 = 0

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

Now, between -4 and 1 which is greater? f(x) or g(x)?

(x + 4)(x - 1) = g(x)-f(x)

In (-4, 1):

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

(x - 1) < 1 - 1 = 0

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

f(x)>g(x)

Let's integrate f(x)-g(x)

The antiderivative of
f(x)-g(x) = 4 - x^2-3x is 4x-(x^3)/3-3(x^2)/2 + c

Put x=1
(*) 4 - (1^3)/3 - 3(1^2)/2 = 4 - 1/3 -3/2 = 4 - (2/6 + 9/6) =
= 4 - 11/6 = 4 - 1 - 5/6 = 2+1/6

Put x=-4
(**) 4*(-4) - (-4^3)/3 -3[(-4)^2]/2 = -16 + 64/3 - 48/2 =
= -16 + 21 + 1/3 - 24 = -19 + 1/3

Subtracting (**) from (*)

2 + 1/6 -(-19 + 1/3) = 2 + 1/6 + 19 - 1/3 = 21 - 1/6 = 20 + 5/6

Answer 3

first find the points on intersection of the two curves
f(x)=4-x^2 and g(x)=3x
4-x^2=3x then by factoring
(x-1)(x+4)=0 therefore x=1,-4 the values of x
and substitutes the values of to get the ordinate
ordinate = 3, -12
it means that the two curves meet at (1,3) and (-4,-12)
to find the area using integral, use vertical strip
where y = f(x)-g(x) with limits from x=-4 to x=1
the integral of (4-x^2-3x)dx with limits x=-4 to x=1
=4x-1/3(x)^3-3/2(x)^2 limits x=-4 to x=1
substitute the limits:
=4[(1)-(-4)]-1/3[(1)^3-(-4)^3]-2/3[(1)^2-(-4)^2]
=125/6 square units