What is area y=3sqrt(x),x+y=4,y=1,y=0 ?

The answer to area y=3sqrt(x),x+y=4,y=1,y=0 is No Solution

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area y=3sqrt(x),x+y=4,y=1,y=0

area

y = 3 x, x + y = 4, y = 1, y = 0

No Solution

Solution steps

Isolate

y

for

x + y = 4 : y = 4 - x

f_{1} (x) = 3 x

f_{2} (x) = 4 - x

To find the intersection points solve

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

3 x = 4 - x : x = 1

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

3 x = 0 : x = 0

4 - x = 1 : x = 3

4 - x = 0 : x = 4

1 = 0 :

False

Curves not bounded

= No Solution

\int \frac{3}{x ^{2} + 3} d x

\frac{d y}{d x} = x^{4} - 1

g (x) = 1 + 6 x + 3 x^{2}

\int \frac{x}{x ^{2} + 4 x + 20} d x

\frac{\partial}{\partial y} (cos (x) cos (y))

What is area y=3sqrt(x),x+y=4,y=1,y=0 ?
The answer to area y=3sqrt(x),x+y=4,y=1,y=0 is No Solution