What is the integral from 1 to 3 of (8x^3+6x-45) ?

The integral from 1 to 3 of (8x^3+6x-45) is 94

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 3 of (8x^3+6x-45)

\int_{1}^{3} (8 x^{3} + 6 x - 45) d x

94

Solution steps

\int_{1}^{3} 8 x^{3} + 6 x - 45 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}^{3} 8 x^{3} d x + \int_{1}^{3} 6 x d x - \int_{1}^{3} 45 d x

\int_{1}^{3} 8 x^{3} d x = 160

\int_{1}^{3} 6 x d x = 24

\int_{1}^{3} 45 d x = 90

= 160 + 24 - 90

Simplify

= 94

\int_{0}^{1} \frac{2 e ^{3 x}}{7} d x

\int_{0}^{a} (b cos (c x) + d) d x

\int_{2}^{3} \frac{1}{x ^{3} + 2 x} d x

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

\int_{1}^{3} sin (r^{2}) r d r

What is the integral from 1 to 3 of (8x^3+6x-45) ?
The integral from 1 to 3 of (8x^3+6x-45) is 94