What is the integral from 2 to 4 of (x^2-4)^{3/2} ?

The integral from 2 to 4 of (x^2-4)^{3/2} is 6ln(2+3^{1/2})+12*3^{1/2}

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)^{3/2}

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

6 ln (2 + 3^{\frac{1}{2}}) + 12 \cdot 3^{\frac{1}{2}}

Decimal Notation

28.68635 \dots

Solution steps

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

Apply Integration By Parts

: [x (x^{2} - 4)^{\frac{3}{2}} - \int 3 x^{2} (x^{2} - 4)^{\frac{1}{2}} d x]_{2}^{4}

= [x (x^{2} - 4)^{\frac{3}{2}} - \int 3 x^{2} (x^{2} - 4)^{\frac{1}{2}} d x]_{2}^{4}

\int 3 x^{2} (x^{2} - 4)^{\frac{1}{2}} d x = 48 - \frac{1}{32} x (x^{2} - 4)^{\frac{1}{2}} + 4 ln \frac{( x ^{2} - 4 ) ^{\frac{1}{2}} + x}{2} + \frac{1}{64} x^{3} (x^{2} - 4)^{\frac{1}{2}}

= x (x^{2} - 4)^{\frac{3}{2}} - 48 - \frac{1}{32} x (x^{2} - 4)^{\frac{1}{2}} + 4 ln \frac{( x ^{2} - 4 ) ^{\frac{1}{2}} + x}{2} + \frac{1}{64} x^{3} (x^{2} - 4)^{\frac{1}{2}}_{2}^{4}

Compute the boundaries:

6 ln (2 + 3^{\frac{1}{2}}) + 12 \cdot 3^{\frac{1}{2}}

= 6 ln (2 + 3^{\frac{1}{2}}) + 12 \cdot 3^{\frac{1}{2}}

\int_{0}^{t^{4}} u d u

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

\int_{0}^{5} k x^{\frac{3}{2}} d x

\int_{0}^{π} 2 cos (x) d x

\int_{6}^{9} x d x

What is the integral from 2 to 4 of (x^2-4)^{3/2} ?
The integral from 2 to 4 of (x^2-4)^{3/2} is 6ln(2+3^{1/2})+12*3^{1/2}