How do you find the area between y=x, #y=1/x^2#, the xaxis and x=3?

Question

Emma Aldridge · Accepted Answer

#S=7/6#

First we note that:

#1/x^2 > 0# for any #x#

so the intercept between the curves and the #x#-axis must belong to the curve #y=x# and in fact occurs for #x=0#.

Then we analyze the inequality:

#x <= 1/x^2#

#x^3 <= 1#

#x<= 1#

This means that in the interval #[0,1]# the graph of #y=x# is below that of #y=1/x^2# while in the interval #[1,3]# the opposite occurs.

A picture can clarify that based on these considerations the area we seek is:

#S = int_0^1 xdx + int_1^3 (dx)/x^2#

Performing the integral we obtain:

#S = int_0^1 xdx + int_1^3 (dx)/x^2 = [x^2/2]_0^1 -[1/x]_1^3 = 1/2-0-1/3 +1= 7/6 #

Cameron Alexander · Answer

undefined To find the area between the curves $y = x$, $y = \frac{1}{x^2}$, the x-axis, and $x = 3$:

1. Identify the points of intersection by setting the two functions equal to each other and solving for $x$.
2. Determine the limits of integration based on the points of intersection and the boundaries given ($x = 0$ and $x = 3$).
3. Integrate the function that lies above the other with respect to $x$ within the determined limits.
4. Subtract the integral of the function below from the integral of the function above to find the area between the curves.

Adam Adkins · Answer

undefined To find the area between the curves $y = x$, $y = \frac{1}{x^2}$, and the x-axis from $x = 0$ to $x = 3$, you need to calculate the definite integral of the absolute difference between the two functions over the given interval.

First, identify the points of intersection of the curves. Setting $y = x$ equal to $y = \frac{1}{x^2}$, we have:

$$x = \frac{1}{x^2}$$

Solving for $x$ gives:

$$x^3 = 1$$

$$x = 1$$

So, the curves intersect at $x = 1$.

Next, determine which function is above the other on the interval [0, 1] and which one is above on the interval [1, 3].

For $x$ between 0 and 1, $y = x$ is above $y = \frac{1}{x^2}$.

For $x$ between 1 and 3, $y = \frac{1}{x^2}$ is above $y = x$.

Now, set up the definite integral:

$$Area = \int_{0}^{1} (x - \frac{1}{x^2}) \, dx + \int_{1}^{3} (\frac{1}{x^2} - x) \, dx$$

Integrate each part separately:

$$Area = \left[\frac{x^2}{2} + \frac{1}{x}\right]_{0}^{1} + \left[-\frac{1}{x} - \frac{x^2}{2}\right]_{1}^{3}$$

$$Area = \left(\frac{1}{2} + 1\right) - \left(0 + \frac{1}{0}\right) + \left(-\frac{1}{3} - \frac{9}{2}\right) - \left(-1 - \frac{1}{2}\right)$$

$$Area = \frac{3}{2} - \frac{1}{3} - \frac{9}{2} + 1 + \frac{1}{2}$$

$$Area = \frac{1}{2} - \frac{8}{2} + 1 + \frac{1}{2}$$

$$Area = \frac{2}{2}$$

$$Area = 1$$

So, the area between the curves $y = x$, $y = \frac{1}{x^2}$, the x-axis, and $x = 3$ is 1 square unit.