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

The answer to area y=3sqrt(x),x+y=4,y=1y=0 is 125/2

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

area

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

\frac{125}{2}

Decimal Notation

62.5

Solution steps

Isolate

x

for

y = 3 x : x = \frac{y ^{2}}{9} {y \geq 0}

Isolate

x

for

x + y = 4 : x = 4 - y

Apply the area formula:

\int_{- 12}^{3} \frac{y ^{2}}{9} - (4 - y) d y

Solve

\int_{- 12}^{3} \frac{y ^{2}}{9} - (4 - y) d y : \frac{125}{2}

The area is:

= \frac{125}{2}

\frac{d}{d x} (2 x^{\frac{3}{2}} + 4 x^{\frac{5}{4}})

\frac{d}{d x} (2 x 25 - x^{2})

\int (x^{5} e^{x^{6}}) d x

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

\frac{\partial}{\partial x} (9 x)

What is area y=3sqrt(x),x+y=4,y=1y=0 ?
The answer to area y=3sqrt(x),x+y=4,y=1y=0 is 125/2