What is the integral from 0 to x^3 of f(t) ?

The integral from 0 to x^3 of f(t) is (x^6f)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to x^3 of f(t)

\int_{0}^{x^{3}} f (t) d t

\frac{x ^{6} f}{2}

Solution steps

\int_{0}^{x^{3}} f t d t

Take the constant out:

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

= f \cdot \int_{0}^{x^{3}} t d t

Apply the Power Rule

: [\frac{t ^{2}}{2}]_{0}^{x^{3}}

= f [\frac{t ^{2}}{2}]_{0}^{x^{3}}

Compute the boundaries:

\frac{x ^{6}}{2}

= f \frac{x ^{6}}{2}

Simplify

= \frac{x ^{6} f}{2}

\int_{1}^{4} 5 x d x

\int_{2}^{9} (3 x + 5) d x

\int_{0}^{x} (cos (x) - sin (y)) d y

\int_{- 1}^{l n (2)} \frac{3 t}{e ^{t}} d t

\int_{\frac{1}{2}}^{1} (x^{- 3} - 22) d x

What is the integral from 0 to x^3 of f(t) ?
The integral from 0 to x^3 of f(t) is (x^6f)/2