Area between two curves?

Question

a) calculate [-2,2] integral x^5
b) Why is the answer the way it is?
c) find the "actual" area between the curve and the x-axis from x=-2 and x=2

Answer 1

a) integral x^5 = (x^6)/6 +c
this you have to split it into to integrals
[-2,0] and [0,2]
or it won't work.
plug it in,evaluate.
answer = 64/3

b)you means hows its calculated?
you should know integral x^5 = (x^6)/6 +c
plug (2^6)/6+c -[(0^6)/6 +c] = 32/3
do it for [-2,0] add it = answer

c) integral is the actual area under the curve.

Answer 2

a)

Integral (-2 to 2, x^5)dx =

(1/6) (x^6) {evaluated from -2 to 2}
(1/6) (2)^6 - (1/6) (-2)^6 = (1/6)2^6 - (1/6)2^6 = 0

b) The answer is the way it is because integrals take into account negative area.

c) The actual area is found by noting we are finding the area between two curves. Our two curves are:

f(x) = x^5, g(x) = 0

When calculating the area, it is always "higher curve" minus "lower curve". Since we're given our bounds for integration, we have to determine whether these two curves intersect anywhere in between.

To determine any intersection, all we have to do is equate the two functions, f(x) = g(x), and then solve for x.

x^5 = 0, implies x = 0.

So the two curves intersect at x = 0, which is within the interval [-2, 2].

Now, we determine which is the higher curve from [-2 to 0], and which is the higher curve from [0 to 2]. When I say higher curve, we determine which has the greater value at the stated interval (f(x) vs g(x)).

To determine the higher curve from -2 to 0, test x = -1. We want to compare f(-1) and g(-1). f(-1) = (-1)^5 = -1, and g(-1) = 0.
Clearly, g(x) is higher. So the first part of our integral will be

A1 = Integral (-2 to 0, g(x) - f(x) ) dx

To determine the higher curve from 0 to 2, test x = 1. Then
f(1) = 1^5 = 1, and g(1) = 0. Clearly, f(x) is higher, so

A2 = Integral (0 to 2, f(x) - g(x)) dx

Now, we solve the actual area as the sum of these two areas.

A = Integral (-2 to 0, g(x) - f(x) ) dx + Integral (0 to 2, f(x) - g(x)) dx

Let's solve for these areas individually.

A1 = Integral (-2 to 0, g(x) - f(x) ) dx
A1 = Integral (-2 to 0, 0 - x^5) dx
A1 = Integral (-2 to 0, -x^5) dx

Let's pull out the (-1) as a constant.

A1 = (-1) * Integral (-2 to 0, x^5) dx

Now, we can solve for the integral directly.

A1 = (-1) * [(1/6)x^6] {evaluated from -2 to 0}
A1 = (-1) [(1/6)(0)^6 - (1/6)(-2)^6]
A1 = (-1) [0 - (1/6) (64)]
A1 = 64/6 = 32/3

A2 = Integral (0 to 2, f(x) - g(x)) dx
A2 = Integral (0 to 2, x^5 - 0) dx
A2 = Integral (0 to 2, x^5) dx

A2 = (1/6)x^6 {evaluated from 0 to 2}
A2 = (1/6)(2^6) - (1/6)(0)
A2 = (1/6)(64) - 0
A2 = 64/6 = 32/3

A = A1 + A2
A = 32/3 + 32/3 = 64/3

NOTE: As long as you recognize which is the higher curve, you should never get a negative answer for your area. This method of subtracting the higher curve from the lower curve is useful because it avoids the usage of the absolute value (for negative area).