Find area enclosed by means of definite integration

**astartleddeer** · December 27th 2012, 10:06 AM

Hello,

I'm in minor confusion why my answer is wrong as opposed towards the book's answer.

I'm asked to find the area enclosed by the equations,

$y_{1} = (2x -1)\ and\ y_{2} = (3 + 2x - x^2)$

$So\ the\ point\ (s)\ of\ intersection\ implies: (2x - 1) = (3 + 2x - x^2)$

$\Rightarrow x = \pm2$

If I compute the area under each equation, and also consider a sign test on both integrals seperately, then:-

$A_{a} = \int ^{2}_{-2} (3 + 2x - x^2)\ dx = \left[(3)(2) + 2^2 - \frac{2^3}{3}\right] - \left[(3)(-2) + 2^2 + \frac{(2)^3}{3}\right]$

$A_{a} = \frac{20}{3}\ units$

$\Rightarrow A_{a}\ is\ (+ve)$

and,

$A_{b} = \int ^{2}_{-2} (2x -1)\ dx = \left(2^2 -2\right) - \left(2^2 + 2\right) = (-4)\ units$

$\Rightarrow A_{b}\ is\ (-ve)$

Now if I consider another sign test,

$Area\,A_{c},\ enclosed\ by\ both\ equations\ = A_{a} \pm A_{b} \Rightarrow A_{c} = (+) + (-)\ or\ (+) - (-) \Rightarrow A_{c}\ is\ (-ve)\ or\ (+ve)$

The total area available under either equation, with limit $x = \pm2$ , is $A_a$ . Then the area $A_{c}$ , enclosed by $y_{1}$ and $y_{2}$ will be less than $A_{a}$ . So in order for $A_{b}$ to be subtracted from $A_{a}$ it must have its sign reversed because it's negative.

So,

$A_{c} = A_{a} + A{b} = \frac{20}{3} + (-4) = \frac{8}{3}\ units$

However, the book has answer of $\frac{32}{3}\ units$

Therefore,

$A_{c} = \frac{20}{3} - (-4) = \frac{32}{3}\ units$

The enclosed area $A_{c}$ is greater than $A_a$

I don't understand why my answer is wrong?

**Plato** · December 27th 2012, 10:24 AM

Originally Posted by astartleddeer

I'm asked to find the area enclosed by the equations,
$y_{1} = (2x -1)\ and\ y_{2} = (3 + 2x - x^2)$
$So\ the\ point\ (s)\ of\ intersection\ implies: (2x - 1) = (3 + 2x - x^2)$

The integral should be $\int_{ - 2}^2 {\left[ {\left( {3 + 2x - x^2 } \right) - \left( {2x - 1} \right)} \right]dx}$

Graph those together on the same graph and you will see why.

**astartleddeer** · December 27th 2012, 10:32 AM

I've already done that.

If I can see signs could be an issue I have a habit to treat each integral seperately.

The majority of the area under y1 is negative.

I can't see why it's not immiterial if you choose to substract the areas individually or as a lumped integrand?

**astartleddeer** · December 27th 2012, 10:57 AM

I've treated the problem as a lumped integrand and I've got the area enclosed as $\frac{32}{3}$ units.

But I don't understand how?

The area requested is part of the area under $y_{2}$ but this tells me its greater than $A_{a}$ because for the total area under $y_2$ for $x = \pm2$ is $\frac{20}{3}$ .This is what I don't understand.

**earboth** · December 27th 2012, 10:57 AM

Originally Posted by astartleddeer

I've already done that.

If I can see signs could be an issue I have a habit to treat each integral seperately.

The majority of the area under y1 is negative.

I can't see why it's not immiterial if you choose to substract the areas individually or as a lumped integrand?

1. If you want to calculate areas you have to use the absolute values of functions because an area below the x-axis is negative and the area above the x-axis is positive.

2. Normally you can avoid the use of absolute values if you don't integrate "over"(?) a zero. In your case both the straight line and the parabola have zeros in the given interval. That means you have to split the interval. The easiest way is of course to integrate the difference of functions as described by Plato.

3. In the attached sketch the green area is your first integral, the red area your second.

**Plato** · December 27th 2012, 11:01 AM

Originally Posted by astartleddeer

if i can see signs could be an issue i have a habit to treat each integral seperately.

THAT then is your problem.

It is one integral of the upper curve minus the lower curve.

**astartleddeer** · December 27th 2012, 11:11 AM

I see now

Thanks

**astartleddeer** · December 27th 2012, 11:32 AM

@ earboth

What's the computer package you used to build that graph?

That's another reason why I was going wrong. I didn't realise the area UNDER the x-axis turns itself out onto the outside of the curve.

Imagine drawing a line, from the parabola, vertically all the way down through x =2 and then stop until you reach the y value where x = -2, that's how I've always pictured the area under any -f(x) parabola.

This is why I was making the claim on how the segmented area could be possibly bigger than the area under the parabola.

But it all makes sense now.

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