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Find the area of the region bounded by the curves $y=x^2$ and $y=x$.

Find the area of the region bounded by the curves $y=x^2+1$ and $y=2$

I have a ton of questions like this and I have been graphing them and then splitting them into intervals and adding them up but this is giving me an answer thats a little off and its taking forever....is there a faster way? Also I am stuck on $y=x^2+1$ and $y=2$ because I dont know what region they want..I see $y=2$ as a line intersecting $x^2+1$, when I graph it.

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3 Answers 3

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HINT They ask for the area of the yellow region: enter image description here

The areas would be given by integrals $\int_{x_1}^{x_2} \left(y_\text{top}(x) - y_\text{bottom}(x)\right) \mathrm{d} x$ with appropriate choices of boundaries $x_1$ and $x_2$ and functions $y_\text{top}(x)$ and $y_\text{bottom}(x)$.

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    $\begingroup$ your curves are nice, what software do you use? $\endgroup$
    – NoChance
    Nov 30, 2011 at 21:12
  • $\begingroup$ @EmmadKareem I am using Mathematica (site). $\endgroup$
    – Sasha
    Nov 30, 2011 at 21:13
  • $\begingroup$ pretty nice, thanks. $\endgroup$
    – NoChance
    Nov 30, 2011 at 21:16
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First we will find the area of the region bounded by the curves:

$y = x^2$ ... (i)

and $y = x $ ... (ii)

To determine the shaded area between these two curves, we need to sketch these curves on a graph. enter image description here

Now, we will find the area of the shaded region from O to A.

Area of Shaded Region Between Two Curves :

$A = \displaystyle \int _a^b [f(x)-g(x)] \;dx$

Where, $f(x)$ is the top curve

$g(x)$ is the bottom curve

$a$ (Lower limit) = $x$ coordinate of extreme left intersection point of the area to be found.

$b$ (Upper limit) = $x$ coordinate of extreme right intersection point of the area to be found.

So, $f(x) = y = x$

$g (x) = y = x^2$

We need to find the limits, $a$ and $b$.

How to find the limits ?

Since limits, $a$ and $b$, are the $x$ coordinates of the intersection points, So, we will find the intersection points of the given curves.

Put the value of $y$ from equation (ii) into equation (i)

$x=x^2$

$x^2-x=0$

$x(x-1)=0$

$ x=0, x = 1 $

Put these values in equation (ii)

$y = 0, \; y = 1$

Thus, the points of intersection are $O(0,0)$ and $A (1,1)$

$\therefore \;a=0, \;b=1$

Area between Curves :

The are will be, $A = \displaystyle\int _a^b [f(x)-g(x)] dx$

$A=\displaystyle \int_0^1\; (x-x^2)\;dx$

$A=\displaystyle \int_0^1\; x\;dx-\displaystyle \int_0^1\; x^2\;dx$

$ =\left ( \dfrac {x^2}{2}\right)_0^1-\left ( \dfrac {x^3}{3}\right)_0^1 $

On putting limits,

$ =\left ( \dfrac {1}{2}-0\right)-\left ( \dfrac {1}{3}-0\right) $

$=\dfrac {1}{2}-\dfrac {1}{3}$

$A=\dfrac {1}{6}$

(II) Now, we will find the shaded area bounded by the curves:

$y = x ^2 + 1$ ... (iii)

$y = 2$ ... (iv)

Curves on Graph : enter image description here

We will find the area of the shaded region from $A$ to $B$

Here, $ f(x) = y = 2$

$g(x) = y = x^2+1 $

Finding the limits using intersection points :

Put the value of $y$ in equation (iii)

$2 = x^2+1 $

$ x^2 = 1 $

$ x = -1, x = 1$

Put these values in equation (iii)

$y = 2, y = 2 $

Thus, the points of intersection are $A (-1, 2)$ and $B (1, 2)$

$ \therefore \;a=-1, \;b=1$

Area between Curves :

The area will be,

$A = \displaystyle\int _a^b [f(x)-g(x)]dx $

$A=\displaystyle \int_{-1}^{1}\; [2-(x^2+1)]\;dx$

$A=\displaystyle \int_{-1}^{1}\; 1\;dx-\displaystyle \int_{-1}^{1}\; x^2\;dx$

$=\left ( x \right)_{-1}^{1}-\left ( \dfrac {x^3}{3}\right)_{-1}^{1}$

On putting limits, we get,

$= (1+1)- \left( \dfrac {1}{3}+\dfrac {1}{3}\right)$

$=2-\dfrac {2}{3}$

$A=\dfrac {4}{3}$

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The fastest way to find the area is to use integration. The area is the result of definite integral of the difference between the two functions.

The area bounded by $y=x^2+1$ and $y=2$ is shown below:

enter image description here

It is the area between the curve and the line of course.

First, we have 2 points that we need to find. The points are those of the intersection of the line and the curve. You can find the points by solving the equation:

$y=x^2+1$ and $y=2$.

You get

$y=x^2+1=2$ and this gives you the points:

$(-1,2)$ and $(1,2)$.

The area A is:

$A=\int_{x=-1}^{x=1} 2-(x^2+1) dx =4/3$.

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