# Area of a region II

#### paulmdrdo

##### Active member

again i just want to check if my answers were correct. i didn't include the complete solution because i'm in a hurry our finals is just around the corner. if my answers are not correct just say so and i will try to solve it again until i get it. thanks!

1. find the area bounded by $y^2=4x$ & $y=2x-4$.

points of intersection are x=4,-2

$\displaystyle\int_{-2}^4 (\frac{1}{2}x+2-\frac{1}{4}x^2)dx$ = $\left[\frac{x^2}{4}+2x-\frac{x^3}{12}\right]_{-2}^4$

it turns out that the area is 16 sq. units

2. find the area bounded by $y=x+3$ and $y=x^2+x-13$

points of intersection are x=4,-4

$\displaystyle\int_{-4}^4 (16-x^2)dx = \left[16x-\frac{x^3}{3}\right]_{-4}^4$

the area is 256/3 sq. units.

Last edited:

#### Petrus

##### Well-known member

2. find the area bounded by $y=x+3$ and $y=x^2+x-13$

points of intersection are x=4,-4

$\displaystyle\int_{-4}^4 (16-x^2)dx = \left[16x-\frac{x^3}{3}\right]_{-4}^4$

the area is 256/3 sq. units.

Hello,
I have not check your answer but could you solve 2 in another way( a way that make it more easy to put in the value after you integrate) You should be able to do this.

Tips:
Look what MarkFL comment on your early topic

#### paulmdrdo

##### Active member
i don't know how to do that. but markfl said in my other topic that i should use even-function rule which i have no idea about.

#### Ackbach

##### Indicium Physicus
Staff member
There are a couple of theorems about symmetries of functions and integrals on symmetric regions. If $f$ is an even function (that is, $f(x)=f(-x)$ for all $x$), and suitably well-behaved (continuous is sufficient, but not necessary) then
$$\int_{-a}^{a} f(x) \, dx=2 \int_{0}^{a}f(x) \, dx.$$
If $f$ is an odd function (that is, $f(x)=-f(-x)$ for all $x$), and suitably well-behaved, then
$$\int_{-a}^{a}f(x) \, dx=0.$$

#### paulmdrdo

##### Active member
There are a couple of theorems about symmetries of functions and integrals on symmetric regions. If $f$ is an even function (that is, $f(x)=f(-x)$ for all $x$), and suitably well-behaved (continuous is sufficient, but not necessary) then
$$\int_{-a}^{a} f(x) \, dx=2 \int_{0}^{a}f(x) \, dx.$$
If $f$ is an odd function (that is, $f(x)=-f(-x)$ for all $x$), and suitably well-behaved, then
$$\int_{-a}^{a}f(x) \, dx=0.$$

#### Ackbach

##### Indicium Physicus
Staff member
2 is correct; you could have used the even function trick that Petrus, MarkFL, and I have mentioned to simplify your calculations, but your result is fine.

1 is incorrect. $x=-2$ is not in the domain of $y^{2}=4x$. If you plug in $x=-2$ on the RHS, you get a negative number, whereas the LHS must always be non-negative. I would consider tilting your head on its side, and do a $y$-integral on this one, so that you don't have to break your integral up into two regions.

Start by plotting the region, and see if that gives you an idea.

#### Prove It

##### Well-known member
MHB Math Helper

again i just want to check if my answers were correct. i didn't include the complete solution because i'm in a hurry our finals is just around the corner. if my answers are not correct just say so and i will try to solve it again until i get it. thanks!

1. find the area bounded by $y^2=4x$ & $y=2x-4$.

points of intersection are x=4,-2

$\displaystyle\int_{-2}^4 (\frac{1}{2}x+2-\frac{1}{4}x^2)dx$ = $\left[\frac{x^2}{4}+2x-\frac{x^3}{12}\right]_{-2}^4$

it turns out that the area is 16 sq. units

The points of intersection are where the two functions are equal. We can see from the second that \displaystyle \begin{align*} y = 2x - 4 \implies 2y = 4x - 8 \implies 4x = 2y + 8 \end{align*}, and so substituting into the first equation we get

\displaystyle \begin{align*} y^2 &= 2y + 8 \\ y^2 - 2y - 8 &= 0 \\ ( y - 4 ) ( y + 2) &= 0 \\ y = 4 \textrm{ or } y &= -2 \end{align*}

From here we can see that the two points of intersection are \displaystyle \begin{align*} (x, y) = (4, 4) \end{align*} and \displaystyle \begin{align*} (x , y ) = ( 1, -2) \end{align*}.

Now, if we draw out the region, it would be easiest to evaluate the area of if we dealt with horizontal strips. Each horizontal strip is bounded on the left by the function \displaystyle \begin{align*} x = \frac{1}{4}y^2 \end{align*} and bounded on the right by the function \displaystyle \begin{align*} x = \frac{1}{2}y + 2 \end{align*}. We would then sum these strips between \displaystyle \begin{align*} y = -2 \end{align*} and \displaystyle \begin{align*} y = 4 \end{align*}. So the area can be evaluated as

\displaystyle \begin{align*} A &= \int_{-2}^4{\int_{\frac{1}{4}y^2}^{\frac{1}{2}y + 2}{1\,dx}\,dy} \\ &= \int_{-2}^4{\left[ x \right]_{\frac{1}{4}y^2}^{\frac{1}{2}y + 2}\,dy} \\ &= \int_{-2}^4{ \frac{1}{2}y + 2 - \frac{1}{4}y^2\,dy } \\ &= \left[ \frac{1}{4}y^2 + 2y - \frac{1}{12}y^3 \right] _{-2}^4 \\ &= \left[ \frac{1}{4}(4)^2 + 2(4) - \frac{1}{12}(4)^3 \right] - \left[ \frac{1}{4}(-2)^2 + 2(-2) - \frac{1}{12}(-2)^3 \right] \\ &= \left[ 4 + 8 - \frac{16}{3} \right] - \left[ 1 - 4 + \frac{2}{3} \right] \\ &= 12 - \frac{16}{3} + 3 - \frac{2}{3} \\ &= 15 - \frac{18}{3} \\ &= 15 - 6 \\ &= 9 \end{align*}

So the area is \displaystyle \begin{align*} 9\,\textrm{units}^2 \end{align*}.

#### LATEBLOOMER

##### New member
i assume you switched the variables in 1, but you've done the arithmetic incorrectly. try solving it again and you'll get 9 sq. units.

#### paulmdrdo

##### Active member
why did you multiply the 1st equation by 2? and how can we determine that this technique is valid to a problem?
\displaystyle \displaystyle \begin{align*} y = 2x - 4 \implies 2y = 4x - 8 \implies 4x = 2y + 8 \end{align*}

#### MarkFL

\displaystyle \displaystyle \begin{align*} y = 2x - 4 \implies 2y = 4x - 8 \implies 4x = 2y + 8 \end{align*}
You were given $y^2=4x$, and so multiplying the other equation by $2$ allows $4x$ to be expressed as a function of $y$. If we have two quantities that are equal, can we not multiply both sides by the same value and still have equality?