What is the integral from 1 to 10 of x^2 2/81 (x-1) ?

The integral from 1 to 10 of x^2 2/81 (x-1) is 107/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 10 of x^2 2/81 (x-1)

\int_{1}^{10} x^{2} \frac{2}{81} (x - 1) d x

\frac{107}{2}

Decimal

53.5

Solution steps

\int_{1}^{10} x^{2} \frac{2}{81} (x - 1) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{2}{81} \cdot \int_{1}^{10} x^{2} (x - 1) d x

Expand

x^{2} (x - 1) : x^{3} - x^{2}

= \frac{2}{81} \cdot \int_{1}^{10} x^{3} - x^{2} 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{2}{81} (\int_{1}^{10} x^{3} d x - \int_{1}^{10} x^{2} d x)

\int_{1}^{10} x^{3} d x = \frac{9999}{4}

\int_{1}^{10} x^{2} d x = 333

= \frac{2}{81} (\frac{9999}{4} - 333)

Simplify

\frac{2}{81} (\frac{9999}{4} - 333) : \frac{107}{2}

= \frac{107}{2}

\int_{0}^{b} 8 e^{x} d x

\int_{- 1}^{8} x^{\frac{2}{3}} d x

\int_{0}^{2 π} - 3 sin^{2} (t) + 9 t d t

\int_{0}^{4} \frac{8}{7 x + 6} d x

\int_{- 1}^{1} \frac{x}{∣ x ∣} d x

What is the integral from 1 to 10 of x^2 2/81 (x-1) ?
The integral from 1 to 10 of x^2 2/81 (x-1) is 107/2