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

The integral from-2 to 2 of (1-(x^2)/4)^2 is 32/15

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{32}{15}

Decimal

2.13333 \dots

Solution steps

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

Expand

(1 - \frac{x ^{2}}{4})^{2} : 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{16}

= \int_{- 2}^{2} 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{16} 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_{- 2}^{2} 1 d x - \int_{- 2}^{2} \frac{x ^{2}}{2} d x + \int_{- 2}^{2} \frac{x ^{4}}{16} d x

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

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

\int_{- 2}^{2} \frac{x ^{4}}{16} d x = \frac{4}{5}

= 4 - \frac{8}{3} + \frac{4}{5}

Simplify

4 - \frac{8}{3} + \frac{4}{5} : \frac{32}{15}

= \frac{32}{15}

\int_{0}^{2 x^{2}} 3 x y d y

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

\int_{0}^{3} (3 - x)^{\frac{- 1}{2}} d x

\int_{- 6}^{1} x^{3} d x

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

What is the integral from-2 to 2 of (1-(x^2)/4)^2 ?
The integral from-2 to 2 of (1-(x^2)/4)^2 is 32/15