What is the integral from 0 to 1 of sqrt(36t^2+9t^4) ?

The integral from 0 to 1 of sqrt(36t^2+9t^4) is 5sqrt(5)-8

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 of sqrt(36t^2+9t^4)

\int_{0}^{1} 36 t^{2} + 9 t^{4} d t

55 - 8

Decimal

3.18033 \dots

Solution steps

\int_{0}^{1} 36 t^{2} + 9 t^{4} d t

Factor

36 t^{2} + 9 t^{4} : 3 t^{2} 4 + t^{2}

= \int_{0}^{1} 3 t^{2} 4 + t^{2} d t

Take the constant out:

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

= 3 \cdot \int_{0}^{1} t^{2} 4 + t^{2} d t

t^{2} = (t),

assuming

t \geq 0

= 3 \cdot \int_{0}^{1} t 4 + t^{2} d t

Apply u-substitution

: \int_{2}^{5} u^{2} d u

= 3 \cdot \int_{2}^{5} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{2}^{5}

= 3 [\frac{u ^{3}}{3}]_{2}^{5}

Compute the boundaries:

\frac{5 5}{3} - \frac{8}{3}

= 3 (\frac{5 5}{3} - \frac{8}{3})

Simplify

= 55 - 8

y^{^{'}} = \frac{sin ( 2 x )}{e ^{y}}

\frac{d}{d x} (\frac{e ^{x}}{4 x ^{2} + 1})

\int e^{x} cos (4 y) d x

x \to 1 lim (ln (x - 1))

\int \frac{( x + 1 ) ^{2}}{x ^{3} + x} d x

What is the integral from 0 to 1 of sqrt(36t^2+9t^4) ?
The integral from 0 to 1 of sqrt(36t^2+9t^4) is 5sqrt(5)-8