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Popular >

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

Solution

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

Solution

\frac{8}{3}

+1

Decimal Notation

2.66666 \dots

Solution steps

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

Expand

(- x + 2)^{2} : x^{2} - 4 x + 4

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

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

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

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

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

Simplify

= \frac{8}{3}

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x \to - \infty lim (\frac{1}{4 x})

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\frac{\partial}{\partial y} (e^{z})

Frequently Asked Questions (FAQ)

What is the integral from 0 to 2 of (-x+2)^2 ?
The integral from 0 to 2 of (-x+2)^2 is 8/3