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Popular Calculus >

integral from 0 to 1 of (2-x)^2

Solution

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

Solution

\frac{7}{3}

+1

Decimal

2.33333 \dots

Solution steps

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

Expand

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

= \int_{0}^{1} 4 - 4 x + x^{2} 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}^{1} 4 d x - \int_{0}^{1} 4 x d x + \int_{0}^{1} x^{2} d x

\int_{0}^{1} 4 d x = 4

\int_{0}^{1} 4 x d x = 2

\int_{0}^{1} x^{2} d x = \frac{1}{3}

= 4 - 2 + \frac{1}{3}

Simplify

4 - 2 + \frac{1}{3} : \frac{7}{3}

= \frac{7}{3}

Graph

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Frequently Asked Questions (FAQ)

What is the integral from 0 to 1 of (2-x)^2 ?
The integral from 0 to 1 of (2-x)^2 is 7/3