Math / Calculus

Questionnodule 11 HW : Problem 4 point)
Write an integral in the form $A=\int_{a}^{b} g(x) d x$ such that $A$ expresses the area of the region between the functions $y=\frac{1}{5 x}, y=\frac{1}{x^{2}}$ and $x=11$ . Evaluate the integral to find the area of the region. Alfiswers should be exact. As a suggestion, graph the equations so you can see the region for which you are solving for the area. $a($ lower limit of integration $)=$ $\square$ $b$ (upper limit of integration) $=$ $\square$ $g(x)=$ $\square$ Area of region $A=$ $\square$
Note: You can earn partial credit on this problem.

Studdy Solution

STEP 1

1. The functions $y = \frac{1}{5x}$ and $y = \frac{1}{x^2}$ intersect at some point, defining the limits of integration.
2. The vertical line $x = 11$ is the upper boundary for the region of interest.
3. The area is calculated between the two functions from the point of intersection to $x = 11$ .

STEP 2

1. Find the intersection point of the two functions.
2. Determine the limits of integration.
3. Set up the integral for the area.
4. Evaluate the integral to find the exact area.

STEP 3

Find the intersection point of the two functions by setting them equal:
$\frac{1}{5x} = \frac{1}{x^2}$
Solve for $x$ :
$x^2 = 5x$ $x(x - 5) = 0$
The solutions are $x = 0$ and $x = 5$ .

STEP 4

Determine the limits of integration. The region of interest is between $x = 5$ and $x = 11$ .
Lower limit $a = 5$ .
Upper limit $b = 11$ .

STEP 5

Set up the integral for the area. The function $g(x)$ is the difference between the two functions:
$g(x) = \frac{1}{5x} - \frac{1}{x^2}$
The integral is:
$A = \int_{5}^{11} \left( \frac{1}{5x} - \frac{1}{x^2} \right) \, dx$

STEP 6

Evaluate the integral:
$A = \int_{5}^{11} \left( \frac{1}{5x} - \frac{1}{x^2} \right) \, dx$
Calculate each part separately:
$\int \frac{1}{5x} \, dx = \frac{1}{5} \ln |x|$
$\int \frac{1}{x^2} \, dx = -\frac{1}{x}$
Thus, the integral becomes:
$A = \left[ \frac{1}{5} \ln |x| + \frac{1}{x} \right]_{5}^{11}$
Evaluate at the bounds:
$A = \left( \frac{1}{5} \ln 11 + \frac{1}{11} \right) - \left( \frac{1}{5} \ln 5 + \frac{1}{5} \right)$
Simplify:
$A = \frac{1}{5} (\ln 11 - \ln 5) + \frac{1}{11} - \frac{1}{5}$
$A = \frac{1}{5} \ln \left( \frac{11}{5} \right) + \frac{1}{11} - \frac{1}{5}$
The exact area of the region is:
$\boxed{\frac{1}{5} \ln \left( \frac{11}{5} \right) + \frac{1}{11} - \frac{1}{5}}$

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