Let's break down the problem step by step:
### Part (a)
We need to represent the area under the curve of [tex]\( f(x) = \frac{2}{x} \)[/tex] from [tex]\( x = 1 \)[/tex] to [tex]\( x = b \)[/tex] as an integral.
[tex]\[
\text{Area} = \int_{1}^{b} \frac{2}{x} \, dx
\][/tex]
### Part (b)
We will now evaluate the integral using the Fundamental Theorem of Calculus.
First, we find the indefinite integral of [tex]\( \frac{2}{x} \)[/tex]:
[tex]\[
\int \frac{2}{x} \, dx = 2 \ln|x| + C
\][/tex]
Next, we apply the limits of integration from 1 to [tex]\( b \)[/tex]:
[tex]\[
\text{Area} = \left. 2 \ln|x| \right|_{1}^{b}
\][/tex]
This means we evaluate the antiderivative at [tex]\( x = b \)[/tex] and [tex]\( x = 1 \)[/tex] and subtract the results:
[tex]\[
\text{Area} = 2 \ln|b| - 2 \ln|1|
\][/tex]
Since [tex]\( \ln|1| = 0 \)[/tex], this simplifies to:
[tex]\[
\text{Area} = 2 \ln(b)
\][/tex]
### Part (c)
We need to find the value of [tex]\( b \)[/tex] such that the area is 5.8 square units. Thus, we set up the equation:
[tex]\[
2 \ln(b) = 5.8
\][/tex]
We solve for [tex]\( b \)[/tex] by isolating [tex]\( \ln(b) \)[/tex]:
[tex]\[
\ln(b) = \frac{5.8}{2} = 2.9
\][/tex]
To solve for [tex]\( b \)[/tex], we exponentiate both sides to get rid of the logarithm:
[tex]\[
b = e^{2.9}
\][/tex]
Therefore, the value of [tex]\( b \)[/tex] that makes the area equal to 5.8 square units is:
[tex]\[
b = 18.1741453694431
\][/tex]
In summary:
1. The integral representing the area is [tex]\(\int_{1}^{b} \frac{2}{x} \, dx\)[/tex].
2. The evaluated integral is [tex]\(2 \ln(b)\)[/tex].
3. Solving the equation [tex]\(2 \ln(b) = 5.8\)[/tex] gives us [tex]\(b = e^{2.9} \approx 18.1741453694431\)[/tex].
