What is area y=sqrt(3x),y=2x-18,y=0 ?

The answer to area y=sqrt(3x),y=2x-18,y=0 is 39

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area y=sqrt(3x),y=2x-18,y=0

area

y = 3 x, y = 2 x - 18, y = 0

39

Solution steps

To find the intersection points solve

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

3 x = 2 x - 18 : x = 12

3 x = 0 : x = 0

2 x - 18 = 0 : x = 9

= \int_{0}^{9} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{9}^{12} ∣ f_{2} (x) - f_{1} (x) ∣ d x

= \int_{0}^{9} 3 x - 0 d x + \int_{9}^{12} 2 x - 18 - 3 x d x

\int_{0}^{9} 3 x - 0 d x = 183

\int_{9}^{12} 2 x - 18 - 3 x d x = 39 - 183

= 183 + 39 - 183

Simplify

= 39

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y^{^{''}} + 4 y^{^{'}} = 5

\int \frac{4}{x} + csc (x) cot (x) d x

\frac{d}{d x} (\frac{4 x}{1 + x ^{2}})

What is area y=sqrt(3x),y=2x-18,y=0 ?
The answer to area y=sqrt(3x),y=2x-18,y=0 is 39