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Popular >

integral from 0 to 2 of (-x^2+4)

Solution

\int_{0}^{2} (- x^{2} + 4) d x

Solution

\frac{16}{3}

+1

Decimal

5.33333 \dots

Solution steps

\int_{0}^{2} - x^{2} + 4 d x

Apply the Sum Rule:

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

= - \int_{0}^{2} x^{2} d x + \int_{0}^{2} 4 d x

\int_{0}^{2} x^{2} d x = \frac{8}{3}

\int_{0}^{2} 4 d x = 8

= - \frac{8}{3} + 8

Simplify

- \frac{8}{3} + 8 : \frac{16}{3}

= \frac{16}{3}

Graph

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\int_{0}^{2 π} cos^{8} (x) d x

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\int_{0}^{t^{4}} u^{3} d u

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\int_{0}^{3} \frac{x ^{2}}{x ^{3} + 6} d x

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Frequently Asked Questions (FAQ)

What is the integral from 0 to 2 of (-x^2+4) ?
The integral from 0 to 2 of (-x^2+4) is 16/3