What is area y=(40)/x ,y=-9x+39 ?

The answer to area y=(40)/x ,y=-9x+39 is -40(ln(8/3)-ln(5/3))+39/2

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area y=(40)/x ,y=-9x+39

area

y = \frac{40}{x}, y = - 9 x + 39

- 40 (ln (\frac{8}{3}) - ln (\frac{5}{3})) + \frac{39}{2}

Decimal Notation

0.69985 \dots

Solution steps

Apply the area formula:

\int_{\frac{5}{3}}^{\frac{8}{3}} \frac{40}{x} - (- 9 x + 39) d x

Solve

\int_{\frac{5}{3}}^{\frac{8}{3}} \frac{40}{x} - (- 9 x + 39) d x : - 40 (ln (\frac{8}{3}) - ln (\frac{5}{3})) + \frac{39}{2}

The area is:

= - 40 (ln (\frac{8}{3}) - ln (\frac{5}{3})) + \frac{39}{2}

\frac{d}{d x} (6 x - 2 x)

\frac{1}{15 s ^{3}}

\frac{\partial}{\partial x} ((x - cy)^{9} + (x + cy)^{9})

\frac{x ^{2}}{x + 5}

\int_{- 5}^{6} 0

What is area y=(40)/x ,y=-9x+39 ?
The answer to area y=(40)/x ,y=-9x+39 is -40(ln(8/3)-ln(5/3))+39/2