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

The integral from-1 to 1 of 1/2 (6-6x^2)^2 is 96/5

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 1/2 (6-6x^2)^2

\int_{- 1}^{1} \frac{1}{2} (6 - 6 x^{2})^{2} d x

\frac{96}{5}

Decimal Notation

19.2

Solution steps

\int_{- 1}^{1} \frac{1}{2} (6 - 6 x^{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_{- 1}^{1} (6 - 6 x^{2})^{2} d x

Expand

(6 - 6 x^{2})^{2} : 36 - 72 x^{2} + 36 x^{4}

= \frac{1}{2} \cdot \int_{- 1}^{1} 36 - 72 x^{2} + 36 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{1}{2} (\int_{- 1}^{1} 36 d x - \int_{- 1}^{1} 72 x^{2} d x + \int_{- 1}^{1} 36 x^{4} d x)

\int_{- 1}^{1} 36 d x = 72

\int_{- 1}^{1} 72 x^{2} d x = 48

\int_{- 1}^{1} 36 x^{4} d x = \frac{72}{5}

= \frac{1}{2} (72 - 48 + \frac{72}{5})

Simplify

\frac{1}{2} (72 - 48 + \frac{72}{5}) : \frac{96}{5}

= \frac{96}{5}

\int t (t + 1)^{6} d t

f (x) = x^{2} - 1, at x = 2

y = e^{x}, y = e^{2 x}, x = 0, x = ln (2)

\frac{x - ln ( 1 + x )}{x ^{2}}

f (x) = 4 x^{2} - 4 x + 1, at x = 0

What is the integral from-1 to 1 of 1/2 (6-6x^2)^2 ?
The integral from-1 to 1 of 1/2 (6-6x^2)^2 is 96/5