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

The answer to area y=4x-x^2,y=3-x is -3sqrt(13)+(31sqrt(13))/6

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

area

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

- 313 + \frac{31 13}{6}

Decimal Notation

7.81202 \dots

Solution steps

Apply the area formula:

\int_{\frac{5 - 13}{2}}^{\frac{5 + 13}{2}} 4 x - x^{2} - (3 - x) d x

Solve

\int_{\frac{5 - 13}{2}}^{\frac{5 + 13}{2}} 4 x - x^{2} - (3 - x) d x : - 313 + \frac{31 13}{6}

The area is:

= - 313 + \frac{31 13}{6}

\frac{\partial}{\partial x} (cos (3 x))

\int (sec (x) - \frac{1}{sin ( x )} + cos (x)) d x

\frac{\partial}{\partial x} (sin (\frac{x}{1 + y}))

\int 27 sec^{- 2} (x t a) n^{3} x d x

What is area y=4x-x^2,y=3-x ?
The answer to area y=4x-x^2,y=3-x is -3sqrt(13)+(31sqrt(13))/6