Question

How do you find the area between `f(x)=x^2+2x+1,g(x)=3x+3`?

Answer 1

4.5 square units

Explanation:

The points of intersection of the two functions are at:
`(-1,0)` and `(2,9)`.

So, we have to find the area under the curve of `g(x)`, then subtract the area under the curve of `f(x)` because `g(x)` is above `f(x)` in the interval `[-1,2]`.
`int_-1^2g(x)dx-int_-1^2f(x)dx`
`int_-1^2(3x+3)dx-int_-1^2(x^2+2x+1)dx`

Simplify the are under the curve of the linear equation using geometry, and the area under the curve of the parabola using a fundamental theorem of calculus.

Let `F(x)` be the antiderivative of `f(x)`
`1/2(3)(9)-[F(2)-F(-1)]`
`27/2-[F(2)-F(-1)]`

To simplify this, we need to know `F(x)`. Integrate `f(x)`:
`F(x)=intx^2+2x+1`
`F(x)=1/3x^3+x^2+x+c`
Solve `F(2)` and `F(-1)`:
`F(2)=1/3(2)^3+2^2+2+c=26/3+c`
`F(-1)=1/3(-1)^3+(-1)^2-1+c=-1/3+c`

Now continue to simplify the expression above:
`27/2-[F(2)-F(-1)]`
`13.5-[(26/3cancel(+c))-(-1/3cancel(+c))]`
`=13.5-9`
`=4.5` square units

Answer 2

Start by finding the intersection points of the two curves.

`{(y = x^2 + 2x + 1), (y = 3x + 3):}`

`3x + 3 = x^2 + 2x + 1`

`0 = x^2 - x - 2`

`0 = (x - 2)(x + 1)`

`x = 2 and -1`

`y = 3x + 3`

`y = 9 and 0`

Hence, the points of intersection are `{2, 9}` and `{-1, 0}`

We now do the graph of each function (on the same grid). We find that `g(x)` is above `f(x)` in the area that they share, so we find the area under `g(x)`, and will subtract the area of `f(x)` from that.

`int_(-1)^2 (3x + 3)dx`

`=>3/2x^2 + 3x|_(-1)^2`

`=>3/2(2)^2 + 3(2) - (3/2(-1)^2 + 3(-1))`

`=>3/2(4) + 6 - 3/2 + 3`

`=>6 + 6 - 3/2 + 3`

`=>27/2`

Now for `f(x)`:

`int_(-1)^2 (x^2 + 2x + 1)`

`=>1/3x^3 + x^2 + x|_(-1)^2`

`=>1/3(2)^3 + 2^2 + 2 - (1/3(-1)^3 + (-1)^2 - 1)`

`=> 8/3 + 4 + 2 + 1/3 - 1 + 1`

`=>9`

Now subtract the two areas.

`=>27/2 - 9 = 9/2 = 4.5`

Hence, the area between the graphs of `f(x) = x^2 + 2x + 1` and `g(x) = 3x + 3` is `4.5" units"^2`.

Hopefully this helps!