What is area sqrt(x), 1/3 x,16 ?

The answer to area sqrt(x), 1/3 x,16 is 2971/3

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area sqrt(x), 1/3 x,16

area

x, \frac{1}{3} x, 16

\frac{2971}{3}

Decimal Notation

990.33333 \dots

Solution steps

To find the intersection points solve

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

x = \frac{1}{3} x : x = 9, x = 0

x = 16 : x = 256

\frac{1}{3} x = 16 : x = 48

= \int_{0}^{9} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{9}^{48} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{48}^{256} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{0}^{9} x - \frac{1}{3} x d x + \int_{9}^{48} x - \frac{1}{3} x d x + \int_{48}^{256} x - 16 d x

\int_{0}^{9} x - \frac{1}{3} x d x = \frac{9}{2}

\int_{9}^{48} x - \frac{1}{3} x d x = - 1283 + \frac{777}{2}

\int_{48}^{256} x - 16 d x = 1283 + \frac{1792}{3}

= \frac{9}{2} - 1283 + \frac{777}{2} + 1283 + \frac{1792}{3}

Simplify

= \frac{2971}{3}

y^{^{'}} + \frac{y}{x} = 0

\int \frac{ln ( x + 1 )}{x + 1} d x

\int_{0}^{1} \frac{x ^{3} + 1}{x + 1} d x

ln (\frac{1}{1 + e ^{- x}})

\frac{d}{d x} (3 x e^{- x} - 2 x)