What is the integral from 0 to 2 of-10x^3sqrt(x^2+4) ?

The integral from 0 to 2 of-10x^3sqrt(x^2+4) is -(2(64sqrt(2)+64))/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2 of-10x^3sqrt(x^2+4)

\int_{0}^{2} - 10 x^{3} x^{2} + 4 d x

- \frac{2 ( 64 2 + 64 )}{3}

Decimal Notation

- 103.00644 \dots

Solution steps

\int_{0}^{2} - 10 x^{3} x^{2} + 4 d x

Take the constant out:

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

= - 10 \cdot \int_{0}^{2} x^{3} x^{2} + 4 d x

Apply u-substitution

: \int_{2}^{22} (u^{2} - 4) u^{2} d u

= - 10 \cdot \int_{2}^{22} (u^{2} - 4) u^{2} d u

Expand

(u^{2} - 4) u^{2} : u^{4} - 4 u^{2}

= - 10 \cdot \int_{2}^{22} u^{4} - 4 u^{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - 10 (\int_{2}^{22} u^{4} d u - \int_{2}^{22} 4 u^{2} d u)

\int_{2}^{22} u^{4} d u = \frac{128 2 - 32}{5}

\int_{2}^{22} 4 u^{2} d u = \frac{4 ( 16 2 - 8 )}{3}

= - 10 (\frac{128 2 - 32}{5} - \frac{4 ( 16 2 - 8 )}{3})

Simplify

- 10 (\frac{128 2 - 32}{5} - \frac{4 ( 16 2 - 8 )}{3}) : - \frac{2 ( 64 2 + 64 )}{3}

= - \frac{2 ( 64 2 + 64 )}{3}

6 t \frac{d y}{d t} + y = t^{7}

\frac{2}{x + 2}

f (x) = (- 4 x^{2} + 16)^{2}

\int_{0}^{\frac{π}{4}} ln (sec (x)) d x

\int \frac{8 m ^{2}}{( m ^{3} + 2 ) ^{3}} d m

What is the integral from 0 to 2 of-10x^3sqrt(x^2+4) ?
The integral from 0 to 2 of-10x^3sqrt(x^2+4) is -(2(64sqrt(2)+64))/3