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Popular >

area 6/x ,3x, x/3

Solution

area

\frac{6}{x}, 3 x, \frac{x}{3}

Solution

12 ln (3)

+1

Decimal Notation

13.18334 \dots

Solution steps

To find the intersection points solve

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

\frac{6}{x} = 3 x : x = 2, x = - 2

\frac{6}{x} = \frac{x}{3} : x = 32, x = - 32

3 x = \frac{x}{3} : x = 0

= \int_{- 32}^{- 2} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- 2}^{0} ∣ f_{3} (x) - f_{2} (x) ∣ d x + \int_{0}^{2} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{2}^{32} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- 32}^{- 2} \frac{6}{x} - \frac{x}{3} d x + \int_{- 2}^{0} \frac{x}{3} - 3 x d x + \int_{0}^{2} 3 x - \frac{x}{3} d x + \int_{2}^{32} \frac{6}{x} - \frac{x}{3} d x

\int_{- 32}^{- 2} \frac{6}{x} - \frac{x}{3} d x = 6 ln (3) - \frac{8}{3}

\int_{- 2}^{0} \frac{x}{3} - 3 x d x = \frac{8}{3}

\int_{0}^{2} 3 x - \frac{x}{3} d x = \frac{8}{3}

\int_{2}^{32} \frac{6}{x} - \frac{x}{3} d x = 6 ln (3) - \frac{8}{3}

= 6 ln (3) - \frac{8}{3} + \frac{8}{3} + \frac{8}{3} + 6 ln (3) - \frac{8}{3}

Simplify

= 12 ln (3)

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\frac{d}{d x} (sec^{3} (2 x))

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\int_{- 7}^{- 5} \frac{3}{( x + 6 ) ^{\frac{8}{11}}} d x

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f (x) = 1 - tan (x), at x = - \frac{π}{4}

integral of 7e^{2x}

\int 7 e^{2 x} d x

(dy)/(dx)=((y-1))/(y+x)

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Frequently Asked Questions (FAQ)

What is area 6/x ,3x, x/3 ?
The answer to area 6/x ,3x, x/3 is 12ln(3)