What is area y= 4/x ,y=16x,y= 1/4 x ?

The answer to area y= 4/x ,y=16x,y= 1/4 x is 24ln(2)

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area y= 4/x ,y=16x,y= 1/4 x

area

y = \frac{4}{x}, y = 16 x, y = \frac{1}{4} x

24 ln (2)

Decimal

16.63553 \dots

Solution steps

To find the intersection points solve

f_{i} (x) = f_{j} (x)

\frac{4}{x} = 16 x : x = \frac{1}{2}, x = - \frac{1}{2}

\frac{4}{x} = \frac{1}{4} x : x = 4, x = - 4

16 x = \frac{1}{4} x : x = 0

= \int_{- 4}^{- \frac{1}{2}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{1}{2}}^{0} ∣ f_{3} (x) - f_{2} (x) ∣ d x + \int_{0}^{\frac{1}{2}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{1}{2}}^{4} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 4}^{- \frac{1}{2}} \frac{4}{x} - \frac{1}{4} x d x + \int_{- \frac{1}{2}}^{0} \frac{1}{4} x - 16 x d x + \int_{0}^{\frac{1}{2}} 16 x - \frac{1}{4} x d x + \int_{\frac{1}{2}}^{4} \frac{4}{x} - \frac{1}{4} x d x

\int_{- 4}^{- \frac{1}{2}} \frac{4}{x} - \frac{1}{4} x d x = 12 ln (2) - \frac{63}{32}

\int_{- \frac{1}{2}}^{0} \frac{1}{4} x - 16 x d x = \frac{63}{32}

\int_{0}^{\frac{1}{2}} 16 x - \frac{1}{4} x d x = \frac{63}{32}

\int_{\frac{1}{2}}^{4} \frac{4}{x} - \frac{1}{4} x d x = 12 ln (2) - \frac{63}{32}

= 12 ln (2) - \frac{63}{32} + \frac{63}{32} + \frac{63}{32} + 12 ln (2) - \frac{63}{32}

Simplify

= 24 ln (2)

\int \frac{1}{2 ^{x} + 1} d x

(x + 19) e^{3 x} + 4

m \to - 1 lim (m^{3} - 1)

f (x) = x^{2} - x^{3}, [- 2, 0]

\frac{d}{d x} (7 x - 3)

What is area y= 4/x ,y=16x,y= 1/4 x ?
The answer to area y= 4/x ,y=16x,y= 1/4 x is 24ln(2)