Question

find the area of the region in the first quadrant bounded by the line ( y = 5x ), the line ( x = 5 ), the curve ( y=\frac{5}{x} ), and the ( x )-axis. the total area of the region is (square). (type an exact answer, using radicals as needed.)

Explanation:

Step1: Find intersection point of \(y = 5x\) and \(y=\frac{5}{x}\)

Set \(5x=\frac{5}{x}\), then \(x^{2}=1\). Since we are in the first - quadrant, \(x = 1\).

Step2: Split the area into two parts

The area \(A\) is the sum of two integrals.
\(A=\int_{0}^{1}\frac{5}{x}dx+\int_{1}^{5}5x dx\)
For \(\int_{0}^{1}\frac{5}{x}dx\), we know that \(\int\frac{1}{x}dx=\ln|x|+C\), so \(\int_{0}^{1}\frac{5}{x}dx=\lim_{a
ightarrow0^{+}}\int_{a}^{1}\frac{5}{x}dx=\lim_{a
ightarrow0^{+}}[5\ln x]_{a}^{1}=\lim_{a
ightarrow0^{+}}(5\ln1 - 5\ln a)=+\infty\) (This is wrong. We should split the area as \(A=\int_{0}^{1}5x dx+\int_{1}^{5}\frac{5}{x}dx\))

For \(\int_{0}^{1}5x dx\), using the power rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), we have \(\int_{0}^{1}5x dx=5\times\frac{x^{2}}{2}\big|_{0}^{1}=\frac{5}{2}(1^{2}-0^{2})=\frac{5}{2}\)

For \(\int_{1}^{5}\frac{5}{x}dx\), using \(\int\frac{1}{x}dx=\ln x + C\), we have \(\int_{1}^{5}\frac{5}{x}dx=5[\ln x]_{1}^{5}=5(\ln5-\ln1)=5\ln5\)

Answer:

\(\frac{5}{2}+5\ln5\)