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

The answer to area y=x,y=x^2+1,x=-1,x=2 is 9/2

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

area

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

\frac{9}{2}

Solution steps

Apply the area formula:

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

Solve

\int_{- 1}^{2} x - (x^{2} + 1) d x : \frac{9}{2}

The area is:

= \frac{9}{2}

y^{^{'}} = \frac{1}{y ^{2}}

y = cos (x), at x = 2 π

\frac{d}{d x} (- 8 sin (8 x))

\frac{d}{d x} ((5 x^{3} + 8 x - 3)^{\frac{3}{2}})

\frac{d}{d x} (2^{x})

What is area y=x,y=x^2+1,x=-1,x=2 ?
The answer to area y=x,y=x^2+1,x=-1,x=2 is 9/2