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

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

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

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

\frac{8}{3}

Decimal Notation

2.66666 \dots

Solution steps

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

(3, - 2), (4, - 7)

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

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

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

y^{^{''}} + 9 y = 30 sin (3 t)

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