What is area x^2-1,-x+2,x=0x=1 ?

The answer to area x^2-1,-x+2,x=0x=1 is 3sqrt(13)-(5sqrt(13))/6

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area x^2-1,-x+2,x=0x=1

area

x^{2} - 1, - x + 2, x = 0 x = 1

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} - 1 - (- x + 2) d x

Solve

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

The area is:

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

x^{4} + 9 e^{x}

\frac{d y}{d x}, y = e^{x}

\frac{d y}{d t} - \frac{2 t}{1 + t ^{2}} y = 3

tan^{2} (y)

\frac{\partial}{\partial x} (2 x^{3} + 3 y^{2} - 5 x y)

What is area x^2-1,-x+2,x=0x=1 ?
The answer to area x^2-1,-x+2,x=0x=1 is 3sqrt(13)-(5sqrt(13))/6