What is the integral from 0 to 1/2 of (4x+1)^3 ?

The integral from 0 to 1/2 of (4x+1)^3 is 5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1/2 of (4x+1)^3

\int_{0}^{\frac{1}{2}} (4 x + 1)^{3} d x

5

Solution steps

\int_{0}^{\frac{1}{2}} (4 x + 1)^{3} d x

Apply u-substitution

: \int_{1}^{3} \frac{u ^{3}}{4} d u

= \int_{1}^{3} \frac{u ^{3}}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{1}^{3} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{1}^{3}

= \frac{1}{4} [\frac{u ^{4}}{4}]_{1}^{3}

Compute the boundaries:

20

= \frac{1}{4} \cdot 20

Simplify

= 5

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

\int_{0}^{1} \frac{sin ( x ^{4} )}{x ^{4}} d x

\int_{0}^{3} (6 x - x^{2}) - x^{2} d x

\int_{- 10}^{10} x d x

\int_{0}^{3} \frac{1}{2} x d x

What is the integral from 0 to 1/2 of (4x+1)^3 ?
The integral from 0 to 1/2 of (4x+1)^3 is 5