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

The integral from 0 to 1 of ((-x+1)^2)/2 is 1/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{6}

Decimal Notation

0.16666 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{1} (- x + 1)^{2} d x

Expand

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

= \frac{1}{2} \cdot \int_{0}^{1} x^{2} - 2 x + 1 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

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

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

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

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

= \frac{1}{2} (\frac{1}{3} - 1 + 1)

Simplify

\frac{1}{2} (\frac{1}{3} - 1 + 1) : \frac{1}{6}

= \frac{1}{6}

\int_{0}^{1} c y^{2} (1 - y)^{4} d y

\int_{0}^{1} \frac{72}{x ^{2} - 81} d x

\int_{- 10}^{10} 29 d t

\int_{0}^{N} x e^{- x} d x

\int_{0}^{π} 4 x sin (x) d x

What is the integral from 0 to 1 of ((-x+1)^2)/2 ?
The integral from 0 to 1 of ((-x+1)^2)/2 is 1/6