Area Between 2 Curves Homework: Calculating the Area between Two Tangent Curves

[squenshl]

Homework Statement

Calculate the area between $y = -x^2+3x+10$ and $y = -x+14$. Note that $y = -x+14$ is the tangent to the curve $y = -x^2+3x+10$ at the point $(2,12)$.

Homework Equations

The Attempt at a Solution

Is it as simple as calculating $\int_{5}^{14} -x+14+x^2-3x-10 = \int_{5}^{14} x^2-4x-10$?

 

[Mark44]

Homework Statement

Calculate the area between $y = -x^2+3x+10$ and $y = -x+14$. Note that $y = -x+14$ is the tangent to the curve $y = -x^2+3x+10$ at the point $(2,12)$.

Homework Equations

The Attempt at a Solution

Is it as simple as calculating $\int_{5}^{14} -x+14+x^2-3x-10 = \int_{5}^{14} x^2-4x-10$?

Where did the 5 come from?

Anyway, your problem is not well-posed, as posted here. Is there some additional information that you left out? The area of the region between the line and the parabola is infinitely large. Did you sketch a graph?

 

[StoneTemplePython]

Homework Statement

Calculate the area between $y = -x^2+3x+10$ and $y = -x+14$. Note that $y = -x+14$ is the tangent to the curve $y = -x^2+3x+10$ at the point $(2,12)$.

Homework Equations

The Attempt at a Solution

Is it as simple as calculating $\int_{5}^{14} -x+14+x^2-3x-10 = \int_{5}^{14} x^2-4x-10$?

it depends on what you want out of this problem.

your problem statement didn't say that the area is to be calculated over $[5, 14]$, and you are missing a dx in those integrals, but otherwise it looks basically ok.

two concerns:

the problem statement asks for

$\int_{a}^{b} \big \vert (-x+14) - (-x^2+3x+10) \big \vert dx$

how do you know that

$\big \vert (-x+14) - (-x^2+3x+10) \big \vert = (-x+14) - (-x^2+3x+10)$

?

Put differently, how do you know you can just get rid of those absolute value signs? In general this would require a justification. I would always start by graphing it / drawing a picture, but some symbolic / mathematical justification is still needed.

you may also want to confirm $14 - 10 \neq -10$

 

[squenshl]

Sorry about the confusion.
The area is between the line and the parabola and above the x axis. My bounds were $5 \leq x \leq 14$ because that's where the line and curve cut the x-axis after drawing it.

 

[Mark44]

Sorry about the confusion.
The area is between the line and the parabola and above the x axis.

That's still an infinite area, unless there are some other bounds. The region between the parabola and the line, and above the x-axis, is in two parts -- the roughly triangular piece to the right of the point where the two figures intersect, and the part to the left of the intersection point. The portion on the left is infinite in area.

 

[squenshl]

That's still an infinite area, unless there are some other bounds. The region between the parabola and the line, and above the x-axis, is in two parts -- the roughly triangular piece to the right of the point where the two figures intersect, and the part to the left of the intersection point. The portion on the left is infinite in area.

 

[StoneTemplePython]

your problem statement still is underspecified. Where did the integration bounds of [5,14] come from? For instance, why not also integrate from $(-\infty, -2)$ and why not [-2,2]?

 

[squenshl]

your problem statement still is underspecified. Where did the integration bounds of [5,14] come from? For instance, why not also integrate from $(-\infty, -2)$ and why not [-2,2]?

Thanks for the help.
I just calculated $\int_{2}^{15} -x+14 \; dx$ and $\int_{2}^{5} -x^2+3x-10 \; dx$ then took the difference to get $\frac{329}{6}$.

 

[Mark44]

I just calculated $\int_{2}^{15} -x+14 \; dx$ and $\int_{2}^{5} -x^2+3x-10 \; dx$ then took the difference to get $\frac{329}{6}$.

There is also the area to the left of the parabola and below the line.

 

[squenshl]

There is also the area to the left of the parabola and below the line.

Not according to the graph above.

 

[Mark44]

Not according to the graph above.

But the graph doesn't agree with your problem statement in post #1.

Calculate the area between $y = -x^2+3x+10$ and $y = -x+14$.

The area between the line and the parabola includes the portion to the left of the parabola.
The graph implies additional restrictions that aren't given in the problem statement -- i.e., that $x \ge 2$ and $y \ge 0$.

So which one is the problem you're working? The verbal description or the graph? They aren't the same.

 

[Mark44]

Where did the integration bounds of [5,14] come from?

I just calculated $\int_{2}^{15} -x+14 \; dx$ and $\int_{2}^{5} -x^2+3x-10 \; dx$ then took the difference to get $\frac{329}{6}$.

The 15 in the first integral will give you a wrong answer. Was that a typo?

 

[squenshl]

Sorry guys.
The question was to find the shared area in the graph & yes that was a typo.
Cheers.

 

[Mark44]

The question was to find the shared area in the graph

Do you mean "shaded" area?