What is the integral from 0 to 5 of (140x^2+9800) ?

The integral from 0 to 5 of (140x^2+9800) is 164500/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 5 of (140x^2+9800)

\int_{0}^{5} (140 x^{2} + 9800) d x

\frac{164500}{3}

Decimal Notation

54833.33333 \dots

Solution steps

\int_{0}^{5} 140 x^{2} + 9800 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_{0}^{5} 140 x^{2} d x + \int_{0}^{5} 9800 d x

\int_{0}^{5} 140 x^{2} d x = \frac{17500}{3}

\int_{0}^{5} 9800 d x = 49000

= \frac{17500}{3} + 49000

Simplify

\frac{17500}{3} + 49000 : \frac{164500}{3}

= \frac{164500}{3}

y = a cos (t) + t^{2} sin (t)

\frac{\partial}{\partial x} ((x + y) e^{x + y})

x \to 7 lim (4)

n = 0 \sum \infty \frac{1}{1 ^{n}}

6 x - x^{2}

What is the integral from 0 to 5 of (140x^2+9800) ?
The integral from 0 to 5 of (140x^2+9800) is 164500/3