area bounded by two curves

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• Aug 19th 2009, 05:28 PM
linearalgebra
area bounded by two curves
Ive found the antiderivative to the two functions and im at the last part, where you have to find the total area. This is what it looks like so far...

[ -(x^3)/3 -(x^2)/2 +2x ] theres a small 1 at the top of the last square bracket and a small -2 at the bottom (interval of [-1,2])

Which numbers do I sub in to get the final answer??
I keep thinking its just the first one, but my answer keeps coming out wrong. I keep getting 7/6, but my textbook says the final answer is 9/2.

Where am I messing up?
• Aug 19th 2009, 05:43 PM
Chris L T521
Quote:

Originally Posted by linearalgebra
Ive found the antiderivative to the two functions and im at the last part, where you have to find the total area. This is what it looks like so far...

[ -(x^3)/3 -(x^2)/2 +2x ] theres a small 1 at the top of the last square bracket and a small -2 at the bottom (interval of [-1,2])

Which numbers do I sub in to get the final answer??
I keep thinking its just the first one, but my answer keeps coming out wrong. I keep getting 7/6, but my textbook says the final answer is 9/2.

Where am I messing up?

What were the two original functions?
• Aug 19th 2009, 06:18 PM
linearalgebra
Quote:

Originally Posted by Chris L T521
What were the two original functions?

y= 2-(x^2)
y=x
• Aug 19th 2009, 06:39 PM
Chris L T521
Quote:

Originally Posted by linearalgebra
y= 2-(x^2)
y=x

First, find the intersection points

$x=2-x^2\implies x^2+x-2=0\implies\left(x+2\right)\left(x-1\right)=0\implies x=1$ or $x=-2$

So, the area is $\int_{-2}^1\left[\left(2-x^2\right)-x\right]\,dx=\int_{-2}^1 2-x-x^2\,dx=\left.\left[2x-\tfrac{1}{2}x^2-\tfrac{1}{3}x^3\right]\right|_{-2}^1$ $=\left(2-\tfrac{1}{2}-\tfrac{1}{3}\right)-\left(-4-2+\tfrac{8}{3}\right)=5-\tfrac{1}{2}=\tfrac{9}{2}$

Does this make sense?
• Aug 19th 2009, 07:08 PM
linearalgebra
Quote:

Originally Posted by Chris L T521
First, find the intersection points

$x=2-x^2\implies x^2+x-2=0\implies\left(x+2\right)\left(x-1\right)=0\implies x=1$ or $x=-2$

So, the area is $\int_{-2}^1\left[\left(2-x^2\right)-x\right]\,dx=\int_{-2}^1 2-x-x^2\,dx=\left.\left[2x-\tfrac{1}{2}x^2-\tfrac{1}{3}x^3\right]\right|_{-2}^1$ $=\left(2-\tfrac{1}{2}-\tfrac{1}{3}\right)-\left(-4-2+\tfrac{8}{3}\right)=5-\tfrac{1}{2}=\tfrac{9}{2}$

Does this make sense?

The last bit, right before you give your final answer..you have two parts. the first part is where you sub in x=1 and then you subtract the second part in which you sub in x=-2. am i correct?

If so, then thank you SO much, you made everything so much clearer
• Aug 19th 2009, 07:25 PM
Chris L T521
Quote:

Originally Posted by linearalgebra
The last bit, right before you give your final answer..you have two parts. the first part is where you sub in x=1 and then you subtract the second part in which you sub in x=-2. am i correct?

If so, then thank you SO much, you made everything so much clearer

Yes, that's correct. Its a direct application of the Fundamental Theorem of Calculus (Pt. II):

$\int_a^bf\!\left(x\right)\,dx=F\!\left(b\right)-F\!\left(a\right)$.
• Aug 19th 2009, 07:29 PM
linearalgebra
awesome. thanks again!