Question

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

Answer 1

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

Explanation:

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`