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Popular Calculus >

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

Solution

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

Solution

\frac{1}{3}

+1

Decimal

0.33333 \dots

Solution steps

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

Expand

(1 - y)^{2} : 1 - 2 y + y^{2}

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

\int_{0}^{1} 1 d y = 1

\int_{0}^{1} 2 y d y = 1

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

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

Simplify

= \frac{1}{3}

Graph

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Frequently Asked Questions (FAQ)

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