What is area y=-x+4,y=x^2+3x ?

The answer to area y=-x+4,y=x^2+3x is -(80sqrt(2))/3+48sqrt(2)

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area y=-x+4,y=x^2+3x

area

y = - x + 4, y = x^{2} + 3 x

- \frac{80 2}{3} + 482

Decimal Notation

30.16988 \dots

Solution steps

Apply the area formula:

\int_{- 2 (1 + 2)}^{2 (2 - 1)} - x + 4 - (x^{2} + 3 x) d x

Solve

\int_{- 2 (1 + 2)}^{2 (2 - 1)} - x + 4 - (x^{2} + 3 x) d x : - \frac{80 2}{3} + 482

The area is:

= - \frac{80 2}{3} + 482

\frac{d}{d x} (x e^{- 4 x})

\frac{\partial}{\partial x} (arctan (y^{3} x))

\int \frac{x}{9 + x ^{2}} d x

\frac{d}{d x} (\frac{x ^{2} + 3}{x ^{2} + 1})

\frac{\partial}{\partial y} (\frac{x ^{2}}{4 - y ^{2}})

What is area y=-x+4,y=x^2+3x ?
The answer to area y=-x+4,y=x^2+3x is -(80sqrt(2))/3+48sqrt(2)