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

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)

\int_{- 5}^{6} 0

\frac{d}{d x} (- 6 sin (x) - 6 x cos (x) + x^{2} sin (x))

d er i v a t i v e y = x^{8} f (x)

x \to \infty lim (\frac{5 x}{5 x + 3 x ^{\frac{1}{x}}})

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