# Area between curve help?

• Apr 13th 2009, 01:58 PM
GlobalCooling
Area between curve help?
Let R be the region in the first quadrant bounded above by y=4x+3 and y=x^2+3

A. Find the area of R

I did this part and I got 10.6667. Can anyone confirm this?

B. The line x=c divides R into two regions. If the area of region R to the left of x=c is (2/3) the area of the region, find the value of c.

I need help with this.

C. The line y=a divides R into two regions of equal area. Find the value of a.

Thank you all!
• Apr 13th 2009, 02:17 PM
icemanfan
Quote:

Originally Posted by GlobalCooling
Let R be the region in the first quadrant bounded above by y=4x+3 and y=x^2+3

A. Find the area of R

I did this part and I got 10.6667. Can anyone confirm this?

B. The line x=c divides R into two regions. If the area of region R to the left of x=c is (2/3) the area of the region, find the value of c.

I need help with this.

C. The line y=a divides R into two regions of equal area. Find the value of a.

Thank you all!

For B, you want to find c such that $\int_0 ^c (4x - x^2) dx = \frac{2}{3} \cdot \frac{32}{3} = \frac{64}{9}$.

For C, you need to calculate the value of four integrals:

$\int_0 ^{\frac{a}{4}} (4x - x^2) dx$

$\int_{\frac{a}{4}} ^{\sqrt{a}} (a - x^2) dx$

$\int_{\frac{a}{4}} ^{\sqrt{a}} (4x - a) dx$

$\int_{\sqrt{a}} ^4 (4x - x^2) dx$

Find the value of a such that:

$\int_0 ^{\frac{a}{4}} (4x - x^2) dx + \int_{\frac{a}{4}} ^{\sqrt{a}} (a - x^2) dx = \int_{\frac{a}{4}} ^{\sqrt{a}} (4x - a) dx + \int_{\sqrt{a}} ^4 (4x - x^2) dx$

Note: I simplified the problem by shifting the graph down 3 units, so you'll have to add 3 to the value of a in order to get the actual answer.
• Apr 13th 2009, 02:25 PM
GlobalCooling
Well I understand what to do for part B, I just don't know how to go about doing it (other than guess and check of course). For part C, would it be easier to do things with respect to y?
• Apr 13th 2009, 02:30 PM
icemanfan
Quote:

Originally Posted by GlobalCooling
Well I understand what to do for part B, I just don't know how to go about doing it (other than guess and check of course). For part C, would it be easier to do things with respect to y?

For part B, evaluate the integral $\int_0 ^c (4x - x^2) dx$.
You should get $2c^2 - \frac{1}{3}c^3$. You know this has to be equal to 2/3 of the area, which is 64/9. All you have to do is solve for c.
• Apr 13th 2009, 02:30 PM
Reckoner
Quote:

Originally Posted by GlobalCooling
Let R be the region in the first quadrant bounded above by y=4x+3 and y=x^2+3

A. Find the area of R

I did this part and I got 10.6667. Can anyone confirm this?

Were you asked for a numerical approximation? If not, you should give an exact answer.

The points of intersection are

$4x+3=x^2+3\Rightarrow x^2-4x=0\Rightarrow x=0,\,4.$

On this interval, $4x+3\geq x^2+3,$ so you set up your integral as

$\int_0^4\!\left[(4x+3)-(x^2+3)\right]dx$

$=\int_0^4(4x-x^2)\,dx$

You shouldn't have any trouble integrating this; leave the answer in fraction form, or at the very least write something like $10.\overline6$ or $10.666\dots$ to indicate that the decimal is periodic and nonterminating.