What is the integral from-1 to 1 of (0-y^2+1) ?

The integral from-1 to 1 of (0-y^2+1) is 4/3

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integral from-1 to 1 of (0-y^2+1)

\int_{- 1}^{1} (0 - y^{2} + 1) d y

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

\int_{- 1}^{1} 0 - y^{2} + 1 d y

Simplify

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

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

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

= - \frac{2}{3} + 2

Simplify

- \frac{2}{3} + 2 : \frac{4}{3}

= \frac{4}{3}

12 y^{^{''}} - 12 y^{^{'}} + 30 y = 0

\int \frac{3 x ^{3}}{4 + x ^{8}} d x

\frac{d x}{d y} = 3 x^{2} y^{3}

\int 3 x^{3} - 4 x^{2} + \frac{7}{x} - \frac{1}{x ^{2}} + 2 e^{x} d x

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{3 x} (x + 1)

What is the integral from-1 to 1 of (0-y^2+1) ?
The integral from-1 to 1 of (0-y^2+1) is 4/3