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

The integral from 2 to 4 of (x^2+4)^3x is 19488

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

19488

Solution steps

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

Apply u-substitution

: \int_{8}^{20} \frac{u ^{3}}{2} d u

= \int_{8}^{20} \frac{u ^{3}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{8}^{20} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{8}^{20}

= \frac{1}{2} [\frac{u ^{4}}{4}]_{8}^{20}

Compute the boundaries:

38976

= \frac{1}{2} \cdot 38976

Simplify

= 19488

\frac{d}{d x} (\frac{ln ( 3 x + 1 )}{6 x + 2})

\frac{9}{sin ( x ) + cos ( x )}

x, 3 x, 4 - x

\int 4 - y^{2} d y

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

What is the integral from 2 to 4 of (x^2+4)^3x ?
The integral from 2 to 4 of (x^2+4)^3x is 19488