What is the integral from-1 to 1 of 3/4 x^2(1-x^2) ?

The integral from-1 to 1 of 3/4 x^2(1-x^2) is 1/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-1 to 1 of 3/4 x^2(1-x^2)

\int_{- 1}^{1} \frac{3}{4} x^{2} (1 - x^{2}) d x

\frac{1}{5}

Decimal Notation

0.2

Solution steps

\int_{- 1}^{1} \frac{3}{4} x^{2} (1 - x^{2}) d x

Take the constant out:

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

= \frac{3}{4} \cdot \int_{- 1}^{1} x^{2} (1 - x^{2}) d x

Expand

x^{2} (1 - x^{2}) : x^{2} - x^{4}

= \frac{3}{4} \cdot \int_{- 1}^{1} x^{2} - 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

= \frac{3}{4} (\int_{- 1}^{1} x^{2} d x - \int_{- 1}^{1} x^{4} d x)

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

\int_{- 1}^{1} x^{4} d x = \frac{2}{5}

= \frac{3}{4} (\frac{2}{3} - \frac{2}{5})

Simplify

\frac{3}{4} (\frac{2}{3} - \frac{2}{5}) : \frac{1}{5}

= \frac{1}{5}

\int - 5 tan^{3} (x) d x

f (3)

y^{^{''}} + y = 9 e^{- 4 x}, y (0) = 0, y^{^{'}} (0) = 0

\frac{d}{d x} (c e^{- 2 x^{2}})

\frac{d}{d x} (sin (x) os (x))

What is the integral from-1 to 1 of 3/4 x^2(1-x^2) ?
The integral from-1 to 1 of 3/4 x^2(1-x^2) is 1/5