What is the integral from 0 to 4 of tsqrt(16-3t) ?

The integral from 0 to 4 of tsqrt(16-3t) is 3008/135

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 4 of tsqrt(16-3t)

\int_{0}^{4} t 16 - 3 t d t

\frac{3008}{135}

Decimal Notation

22.28148 \dots

Solution steps

\int_{0}^{4} t 16 - 3 t d t

Apply u-substitution

: \int_{16}^{4} \frac{u ( u - 16 )}{9} d u

= \int_{16}^{4} \frac{u ( u - 16 )}{9} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{4}^{16} \frac{u ( u - 16 )}{9} d u

Take the constant out:

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

= - \frac{1}{9} \cdot \int_{4}^{16} u (u - 16) d u

Expand

u (u - 16) : u^{\frac{3}{2}} - 16 u

= - \frac{1}{9} \cdot \int_{4}^{16} u^{\frac{3}{2}} - 16 u 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

= - \frac{1}{9} (\int_{4}^{16} u^{\frac{3}{2}} d u - \int_{4}^{16} 16 u d u)

\int_{4}^{16} u^{\frac{3}{2}} d u = \frac{1984}{5}

\int_{4}^{16} 16 u d u = \frac{1792}{3}

= - \frac{1}{9} (\frac{1984}{5} - \frac{1792}{3})

Simplify

- \frac{1}{9} (\frac{1984}{5} - \frac{1792}{3}) : \frac{3008}{135}

= \frac{3008}{135}

\int 3 x^{2} 4 x^{4} d x

x \to 0 lim (4 - 5 x)

f (x) = x^{2} - 5, c = 3

\int_{- 2}^{3} \frac{24}{x ^{4}} d x

x y^{^{'}} - 3 y^{2} = 0, y (e) = 1

What is the integral from 0 to 4 of tsqrt(16-3t) ?
The integral from 0 to 4 of tsqrt(16-3t) is 3008/135