What is the net area between #f(x) = sqrt(x+3)-x^3+6x # and the x-axis over #x in [2, 4 ]#?

Question

Aurora Alvarado · Accepted Answer

#2/3(7sqrt7-5sqrt5-36)approx-19.107#

Note that the area between a curve defined by #f(x)# on the interval #[a,b]# can be expressed as:

#int_a^bf(x)dx#

Thus, the area here is equivalent to:

#int_2^4(sqrt(x+3)-x^3+6x)dx#

The integral can be split up:

#=int_2^4sqrt(x+3)dx-int_2^4x^3dx+6int_2^4xdx#

The second and third terms are more easily integrated: use the rule that #int_a^bx^ndx=[x^(n+1)/(n+1)]_a^b#.

#=int_2^4sqrt(x+3)dx-[x^4/4]_2^4+6[x^2/2]_2^4#

For the first integral, we can use substitution. Let #u=x+3#. Thus, #du=dx#. Don't forget to plug the current bounds into #x+3# given the substitution:

#=int_5^7sqrtudu-[x^4/4]_2^4+6[x^2/2]_2^4#

Integrate again using the aforementioned rule, recalling that #sqrtu=u^(1/2)#:

#=[u^(3/2)/(3/2)]_5^7-[x^4/4]_2^4+6[x^2/2]_2^4#

Bring out multiplicative constants:

#=2/3[u^(3/2)]_5^7-1/4[x^4]_2^4+3[x^2]_2^4#

Evaluate:

#=2/3[7^(3/2)-5^(3/2)]-1/4[4^4-2^4]+3[4^2-2^2]#

#=2/3[7sqrt7-5sqrt5]-1/4[240]+3[12]#

#=2/3[7sqrt7-5sqrt5]-24#

#=2/3(7sqrt7-5sqrt5-36)#

#approx-19.107#

Christopher Allers · Answer

undefined To find the net area between $ f(x) = \sqrt{x+3} - x^3 + 6x $ and the x-axis over $ x $ in the interval $[2, 4]$, follow these steps:

1. Identify the intervals where $ f(x) $ is above the x-axis (positive) and below the x-axis (negative).
2. Integrate the absolute value of $ f(x) $ over these intervals.
3. Subtract the integral of $ f(x) $ where it is below the x-axis from the integral where it is above the x-axis.

The net area can be calculated using definite integrals.

Let's proceed with the calculations:

1. Identify where $ f(x) $ is positive and negative in the interval $[2, 4]$:

Evaluate $ f(x) $ at the interval endpoints:
   - $ f(2) = \sqrt{2 + 3} - 2^3 + 6(2) = \sqrt{5} - 8 + 12 = \sqrt{5} + 4 $
   - $ f(4) = \sqrt{4 + 3} - 4^3 + 6(4) = \sqrt{7} - 64 + 24 = \sqrt{7} - 40 $

Since $ f(2) > 0 $ and $ f(4) < 0 $, $ f(x) $ is positive on the interval $ [2, c] $ and negative on the interval $ [c, 4] $, where $ c $ is the root of $ f(x) $ in the interval $[2, 4]$.

2. Find the root $ c $ of $ f(x) $ in the interval $[2, 4]$:

Set $ f(x) = 0 $ and solve for $ x $:
   $$ \sqrt{x + 3} - x^3 + 6x = 0 $$

There may not be a simple analytical solution to this equation. You can use numerical methods like the bisection method or Newton's method to approximate the root $ c $.

3. Integrate the absolute value of $ f(x) $ over the intervals $ [2, c] $ and $ [c, 4] $:

$$ \text{Net area} = \int_{2}^{c} |f(x)| \, dx - \int_{c}^{4} |f(x)| \, dx $$

Evaluate these integrals numerically once you have found the value of $ c $.

This procedure will give you the net area between $ f(x) $ and the x-axis over the interval $[2, 4]$.