area bounded by two curves

**linearalgebra** · August 19th 2009, 04:28 PM

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?

**Chris L T521** · August 19th 2009, 04:43 PM

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?

**linearalgebra** · August 19th 2009, 05:18 PM

Originally Posted by Chris L T521

What were the two original functions?

y= 2-(x^2)
y=x

**Chris L T521** · August 19th 2009, 05:39 PM

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?

**linearalgebra** · August 19th 2009, 06:08 PM

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

**Chris L T521** · August 19th 2009, 06:25 PM

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):

$\int_a^bf\!\left(x\right)\,dx=F\!\left(b\right)-F\!\left(a\right)$ .

**linearalgebra** · August 19th 2009, 06:29 PM

awesome. thanks again!

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