# Thread: Area enclosed between curves

1. ## Area enclosed between curves

Find the area between the curves 𝑦 = −𝑥 ^2 + 2𝑥 and 𝑦 = 𝑥 − 2 on the interval [0,3].

I seem to be very poor at these in particular ...any advice on how to better them would be helpful

First I set -x^2+2x=x-2
I then did this -X^2+x+2=0
because of the - sign in front of the x^2 I'm having trouble breaking it down ...how ever I wolfram alfa 'd it and the x values become -1 and 2

As it is only looking for the interval between [0 and 3 ]

I then integrated between -1 and 0 giving -13/6
i then integrated between 3 and 0 and got 15/2
or should I integrate between 2 and o giving 2/3

the answer is 31/6 but I can't seem to figure what I must do ....I will practice these types ...hmmm
feel free to unleash your guidance on my faults ...
cheers ...

2. ## Re: Area enclosed between curves

Originally Posted by bee77
Find the area between the curves 𝑦 = −𝑥 ^2 + 2𝑥 and 𝑦 = 𝑥 − 2 on the interval [0,3].

I seem to be very poor at these in particular ...any advice on how to better them would be helpful

First I set -x^2+2x=x-2

I then did this -X^2+x+2=0
because of the - sign in front of the x^2 I'm having trouble breaking it down ...how ever I wolfram alfa 'd it and the x values become -1 and 2
If you are taking calculus, you surely learned to solve quadratic equations long ago. That is the same as (multiply both sides by -1) $x^2- x- 2= 0$ (it isn't necessary to have the coefficient of $x^2$ positive but I prefer it that way).

Now you can factor: since -2= (1)(-2) and -2+ 1= -1 $x^2- x- 2= (x+ 1)(x- 2)= 0$ so either $x+ 1= 0$ and $x= 1$ or $x- 2= 0$ and $x= 2$.

Or you can "complete the square": $x^2- x+ \frac{1}{4}- \frac{1}{4}- 2= \left(x- \frac{1}{2}\right)^2- \frac{9}{4}= 0$ so $\left(x- \frac{1}{2}\right)^2= \frac{9}{4}$, $x- \frac{1}{2}= \pm\frac{3}{2}$ and then $x= \frac{1}{2}+ \frac{3}{2}= 2$ or [tex]x= \frac{1}{2}- \frac{3}{2}= -1[tex].

Or you can use the quadratic formula: $x= \frac{-(-1)\pm\sqrt{(-1)^2- 4(1)(-2)}}{2}= \frac{1\pm\sqrt{9}}{2}= \frac{1\pm 3}{2}$ so $x= \frac{4}{2}= 2$ or $x= \frac{-2}{2}= -1$.

As it is only looking for the interval between [0 and 3 ]

I then integrated between -1 and 0 giving -13/6
Why? As you say above "it is only looking for the interval between [0 and 3 ]" so what happens between -1 and 0 is irrelevant. The only reason to check for the zeroes of the function is to observe that the parabola, $y= -x^2+ 2x$ is above the line $y= x- 2$ for x between 0 and 2 and below it for x between 2 and 3. You can either integrate $(-x^2+ 2x)- (x- 2)$ from 0 to 2 and then integrate $(x- 2)- (-x^2+ 2x)$ from 2 to 3 or (probably simpler) integrate $(-x^2+ 2x)- (x- 2)$ from 0 to 2 and then subtract the integral of $(-x^2+ 2x)- (x- 2)$ from 2 to 3.

i then integrated between 3 and 0 and got 15/2
or should I integrate between 2 and o giving 2/3

the answer is 31/6 but I can't seem to figure what I must do ....I will practice these types ...hmmm
feel free to unleash your guidance on my faults ...
cheers ...

3. ## Re: Area enclosed between curves

Originally Posted by bee77
Find the area between the curves �� = −�� ^2 + 2�� and �� = �� − 2 on the interval [0,3].

I seem to be very poor at these in particular ...any advice on how to better them would be helpful

First I set -x^2+2x=x-2

I then did this -X^2+x+2=0
because of the - sign in front of the x^2 I'm having trouble breaking it down ...how ever I wolfram alfa 'd it and the x values become -1 and 2
If you are taking calculus, you surely learned to solve quadratic equations long ago. That is the same as (multiply both sides by -1) $x^2- x- 2= 0$ (it isn't necessary to have the coefficient of $x^2$ positive but I prefer it that way).

Now you can factor: since -2= (1)(-2) and -2+ 1= -1 $x^2- x- 2= (x+ 1)(x- 2)= 0$ so either $x+ 1= 0$ and $x= 1$ or $x- 2= 0$ and $x= 2$.

Or you can "complete the square": $x^2- x+ \frac{1}{4}- \frac{1}{4}- 2= \left(x- \frac{1}{2}\right)^2- \frac{9}{4}= 0$ so $\left(x- \frac{1}{2}\right)^2= \frac{9}{4}$, $x- \frac{1}{2}= \pm\frac{3}{2}$ and then $x= \frac{1}{2}+ \frac{3}{2}= 2$ or $x= \frac{1}{2}- \frac{3}{2}= -1$.

Or you can use the quadratic formula: $x= \frac{-(-1)\pm\sqrt{(-1)^2- 4(1)(-2)}}{2}= \frac{1\pm\sqrt{9}}{2}= \frac{1\pm 3}{2}$ so $x= \frac{4}{2}= 2$ or $x= \frac{-2}{2}= -1$.

As it is only looking for the interval between [0 and 3 ]

I then integrated between -1 and 0 giving -13/6
Why? As you say above "it is only looking for the interval between [0 and 3 ]" so what happens between -1 and 0 is irrelevant. The only reason to check for the zeroes of the function is to observe that the parabola, $y= -x^2+ 2x$ is above the line $y= x- 2$ for x between 0 and 2 and below it for x between 2 and 3. You can either integrate $(-x^2+ 2x)- (x- 2)$ from 0 to 2 and then integrate $(x- 2)- (-x^2+ 2x)$ from 2 to 3 or (probably simpler) integrate $(-x^2+ 2x)- (x- 2)$ from 0 to 2 and then subtract the integral of $(-x^2+ 2x)- (x- 2)$ from 2 to 3.

i then integrated between 3 and 0 and got 15/2
or should I integrate between 2 and o giving 2/3

the answer is 31/6 but I can't seem to figure what I must do ....I will practice these types ...hmmm
feel free to unleash your guidance on my faults ...
cheers ...

4. ## Re: Area enclosed between curves

wow thanks , yes I've done quadratic equations ...I will carefully go through the problem again and your notes ..the integration between 0 and 2 and then from 2 to 3 is something I'll work on ...So basically when you do find the x values working between the found values ie -1 to 2 but in this case 0 and 2 for the interval [0 and 3] and then 2 and 3 as it was one of the x values within the [0 and 3] interval as well is what I need to do in future similar problems (then subtracting the integrated values for area ) I think i got a bit confused as 3 wasn't one of the x values found but now it makes sense ....hope that makes sense ...thanks a million !!