What is the integral from 1 to 2 of 6/(x^3) ?

The integral from 1 to 2 of 6/(x^3) is 9/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 2 of 6/(x^3)

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

\frac{9}{4}

Decimal Notation

2.25

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

: [- \frac{1}{2 x ^{2}}]_{1}^{2}

= 6 [- \frac{1}{2 x ^{2}}]_{1}^{2}

Compute the boundaries:

\frac{3}{8}

= 6 \cdot \frac{3}{8}

Simplify

= \frac{9}{4}

\int_{0}^{1} \frac{4 x}{π ( 1 + x ^{2} )} d x

\int_{1}^{e^{3}} \frac{1}{t} d t

\int_{- a}^{a} cos^{2} (t) d t

\int_{- 1}^{1} x^{\frac{4}{3}} d x

\int_{0}^{0.5} \frac{1}{x ln ( x )} d x

What is the integral from 1 to 2 of 6/(x^3) ?
The integral from 1 to 2 of 6/(x^3) is 9/4