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

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

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

area

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

313 - \frac{5 13}{6}

Decimal

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 \to - 1 lim (5 - 3 x - x^{2})

(3.1) (0)

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

y = (4 - 3 x^{2})^{3}

y^{^{'}} + 2 y = 10

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