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

The integral from 3 to x^2 of 1/(t^4) is -1/(3x^6)+1/81

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int_{3}^{x^{2}} \frac{1}{t ^{4}} d t

- \frac{1}{3 x ^{6}} + \frac{1}{81}

Solution steps

\int_{3}^{x^{2}} \frac{1}{t ^{4}} d t

Apply the Power Rule

: [- \frac{1}{3 t ^{3}}]_{3}^{x^{2}}

= [- \frac{1}{3 t ^{3}}]_{3}^{x^{2}}

Compute the boundaries:

- \frac{1}{3 x ^{6}} + \frac{1}{81}

= - \frac{1}{3 x ^{6}} + \frac{1}{81}

\int_{- 1}^{1} x^{2} + x + 2 d x

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

\int_{0}^{1} 3.6 x^{3} - 2.4 x^{4} d x

\int_{50}^{100} 7 x d x

\int_{1}^{e} z ln (z) d z

What is the integral from 3 to x^2 of 1/(t^4) ?
The integral from 3 to x^2 of 1/(t^4) is -1/(3x^6)+1/81