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

The integral from 1 to 4 of (3x^2-2)^3 is 1850631/35

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1850631}{35}

Decimal

52875.17142 \dots

Solution steps

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

Expand

(3 x^{2} - 2)^{3} : 27 x^{6} - 54 x^{4} + 36 x^{2} - 8

= \int_{1}^{4} 27 x^{6} - 54 x^{4} + 36 x^{2} - 8 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_{1}^{4} 27 x^{6} d x - \int_{1}^{4} 54 x^{4} d x + \int_{1}^{4} 36 x^{2} d x - \int_{1}^{4} 8 d x

\int_{1}^{4} 27 x^{6} d x = \frac{442341}{7}

\int_{1}^{4} 54 x^{4} d x = \frac{55242}{5}

\int_{1}^{4} 36 x^{2} d x = 756

\int_{1}^{4} 8 d x = 24

= \frac{442341}{7} - \frac{55242}{5} + 756 - 24

Simplify

\frac{442341}{7} - \frac{55242}{5} + 756 - 24 : \frac{1850631}{35}

= \frac{1850631}{35}

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

\int_{0}^{2} (2 x - 9) (8 x^{2} + 2) d x

\int_{1}^{2} \frac{3 - 6 x}{x ^{3}} d x

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

\int_{0}^{2} ∣ x (x - 1) ∣ d x

What is the integral from 1 to 4 of (3x^2-2)^3 ?
The integral from 1 to 4 of (3x^2-2)^3 is 1850631/35