What is area y=-x^2+4,y=1-x ?

The answer to area y=-x^2+4,y=1-x is 3sqrt(13)-(5sqrt(13))/6

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area y=-x^2+4,y=1-x

area

y = - x^{2} + 4, y = 1 - x

313 - \frac{5 13}{6}

Decimal Notation

7.81202 \dots

Solution steps

Apply the area formula:

\int_{- \frac{- 1 + 13}{2}}^{\frac{1 + 13}{2}} - x^{2} + 4 - (1 - x) d x

Solve

\int_{- \frac{- 1 + 13}{2}}^{\frac{1 + 13}{2}} - x^{2} + 4 - (1 - x) d x : 313 - \frac{5 13}{6}

The area is:

= 313 - \frac{5 13}{6}

\int_{1}^{6} \frac{x ^{2} - 1}{x} d x

y^{^{''}} - 2 y^{^{'}} + 2 y = 2 x - 2, y (0) = 0, y (π) = π

\int 3 x^{3} cos (x^{2}) d x

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

x^{2} + y^{2} = 169, (- 5, 12)

What is area y=-x^2+4,y=1-x ?
The answer to area y=-x^2+4,y=1-x is 3sqrt(13)-(5sqrt(13))/6