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

##### 1 Answer

#### Explanation:

Note that the area between a curve defined by

#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_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

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

Integrate again using the aforementioned rule, recalling that

#=[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#