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Popular >

integral from-1 to 1 of-z^2+1

Solution

\int_{- 1}^{1} - z^{2} + 1 d z

Solution

\frac{4}{3}

+1

Solution steps

\int_{- 1}^{1} - z^{2} + 1 d z

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} z^{2} d z + \int_{- 1}^{1} 1 d z

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

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

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

Simplify

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

= \frac{4}{3}

Popular Examples

integral from-pi to pi of sin(θ)

\int_{- π}^{π} sin (θ) d θ

integral from-pi to pi of pi

\int_{- π}^{π} π d x

integral from-1 to 1 of x 1/2

\int_{- 1}^{1} x \frac{1}{2} d x

integral from 0 to 1 of (e^{2x+1})/2

\int_{0}^{1} \frac{e ^{2 x + 1}}{2} d x

integral from 1 to 3 of (x^2+e^{-x})

\int_{1}^{3} (x^{2} + e^{- x}) d x

Frequently Asked Questions (FAQ)

What is the integral from-1 to 1 of-z^2+1 ?
The integral from-1 to 1 of-z^2+1 is 4/3