What is the integral from-1 to 1 of y^2-1-(1-y^2) ?

The integral from-1 to 1 of y^2-1-(1-y^2) is -8/3

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integral from-1 to 1 of y^2-1-(1-y^2)

\int_{- 1}^{1} y^{2} - 1 - (1 - y^{2}) d y

- \frac{8}{3}

Decimal

- 2.66666 \dots

Solution steps

\int_{- 1}^{1} y^{2} - 1 - (1 - y^{2}) d y

Simplify

= \int_{- 1}^{1} 2 y^{2} - 2 d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 1}^{1} 2 y^{2} d y - \int_{- 1}^{1} 2 d y

\int_{- 1}^{1} 2 y^{2} d y = \frac{4}{3}

\int_{- 1}^{1} 2 d y = 4

= \frac{4}{3} - 4

Simplify

\frac{4}{3} - 4 : - \frac{8}{3}

= - \frac{8}{3}

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

x \to - 6 - lim (\frac{( x + 3 )}{x + 6})

\int \frac{1 - x}{x ^{2} - 4} d x

f (x) = x^{2} - \frac{4}{x} - 10, at x = 8

6 x^{2}

What is the integral from-1 to 1 of y^2-1-(1-y^2) ?
The integral from-1 to 1 of y^2-1-(1-y^2) is -8/3