Area between curves

**Open that Hampster!** · November 17th 2009, 04:57 AM

1) $y = x^2-x$
$y = x+3$

Intersects are -1 and 3

$\int_0^1 x^2-x - (x+3)$ = $\int_0^1 x^2-2x-3$

$[\frac{x^3}{3}-x^2-3x]^3_{-1}$

$= (9-9-9)-(-\frac{1}{3}-1+3) = -9+\frac{1}{3}+1-3$ = $-\frac{32}{3}$

2) y = \sqrt{x}
y = \frac{1}{2}x
x = 9

There's an intersection at x = 4, so the Area between curves is:

$\int^4_0\sqrt{x} - \frac{1}{2}x+$ $\int^9_4\sqrt{x} - \frac{1}{2}x$

Shorthanding this answer: the antiderivative is $\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^2}{4}$

And for 4 you get $\frac{16}{3}-4$ , and for 9 you get $\frac{-9}{4}$ and for 0 it's 0, so the answer is:

$\frac{16}{3}-4+[\frac{-9}{4}-\frac{16}{3}-4]$

A bunch of stuff cancels and you get $\frac{-9}{4}$

~~~

The two above I though I did correctly, and didn't apparently, so I don't know where I went wrong on those.

3) Find the number b such that the line y = b divides the region bounded by the curves y = x^2 and y = 4 into two regions with equal area. (Round your answer to the nearest hundredth.)

This one I'm just not sure where to get started.

**RockHard** · November 17th 2009, 06:20 AM

Well obviously your answer for area shouldn't be negative, I suggest drawing the graphs of both if you havent already, usually if you get some sort of negative the first check is to switch the way you subtracted the equation if thats not it, it only leaves two other things your limit of integration are wrong or you integrated wrong

**tonio** · November 17th 2009, 06:25 AM

Originally Posted by Open that Hampster!

1) $y = x^2-x$
$y = x+3$

Intersects are -1 and 3

$\int_0^1 x^2-x - (x+3)$ = $\int_0^1 x^2-2x-3$

$[\frac{x^3}{3}-x^2-3x]^3_{-1}$

$= (9-9-9)-(-\frac{1}{3}-1+3) = -9+\frac{1}{3}+1-3$ = $-\frac{32}{3}$

2) y = \sqrt{x}
y = \frac{1}{2}x
x = 9

There's an intersection at x = 4, so the Area between curves is:

$\int^4_0\sqrt{x} - \frac{1}{2}x+$ $\int^9_4\sqrt{x} - \frac{1}{2}x$

Shorthanding this answer: the antiderivative is $\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^2}{4}$

And for 4 you get $\frac{16}{3}-4$ , and for 9 you get $\frac{-9}{4}$ and for 0 it's 0, so the answer is:

$\frac{16}{3}-4+[\frac{-9}{4}-\frac{16}{3}-4]$

A bunch of stuff cancels and you get $\frac{-9}{4}$

~~~

The two above I though I did correctly, and didn't apparently, so I don't know where I went wrong on those.

3) Find the number b such that the line y = b divides the region bounded by the curves y = x^2 and y = 4 into two regions with equal area. (Round your answer to the nearest hundredth.)

This one I'm just not sure where to get started.

It must be that they mean GEOMETRIC area, as opposed to algebraic area, and thus the result must always be positive, so you have not only to search for the intersection points of the curves but also which function is bigger than (or "above") the other in the integration interval.
For example, in $[-1,3]\,,\,\,g(x)=x+3>x^2-x=f(x)$ and thus you should have integrated $g-f$ and not $f-g$ as you did.
By the way, fix also the lower limit of this integral: it must be -1 and not zero.

As for the second problem: $\sqrt{x}>\frac{1}{2}x\,\,for\,\,0\leq x\leq 4\,,\,\,and\,\,\sqrt{x}<\frac{1}{2}x\,\,for\,\,4<x \leq 9$ , so the wanted area, bearing in mind what we said before, is:

$\int\limits_0^4\left(\sqrt{x}-\frac{1}{2}x\right)dx+\int\limits_4^9\left(\frac{1 }{2}x-\sqrt{x}\right)dx$ .

Where a curve is "above" another is very important in these questions.

Tonio

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