Area of Region II: 16 & 256/3 sq. units

[paulmdrdo1]

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.

 

[Petrus]

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:

Spoiler

Look what MarkFL comment on your early topic

 

[paulmdrdo1]

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]

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
$$ \int_{-a}^{a}f(x) \, dx=0.$$

 

[paulmdrdo1]

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
$$ \int_{-a}^{a}f(x) \, dx=0.$$

are my answers above correct?

 

[Ackbach]

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]

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 $$\displaystyle \begin{align*} 9\,\textrm{units}^2 \end{align*}$$.

 

[paulmdrdo]

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.

 

[paulmdrdo1]

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]

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

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?