What is the integral from 0 to 1 of sqrt(64t^2+9t^4) ?

The integral from 0 to 1 of sqrt(64t^2+9t^4) is (73sqrt(73)-512)/(27)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 1 of sqrt(64t^2+9t^4)

\int_{0}^{1} 64 t^{2} + 9 t^{4} d t

\frac{73 73 - 512}{27}

Decimal

4.13749 \dots

Solution steps

\int_{0}^{1} 64 t^{2} + 9 t^{4} d t

Factor

64 t^{2} + 9 t^{4} : t^{2} 64 + 9 t^{2}

= \int_{0}^{1} t^{2} 64 + 9 t^{2} d t

t^{2} = (t),

assuming

t \geq 0

= \int_{0}^{1} t 64 + 9 t^{2} d t

Apply u-substitution

= \int_{64}^{73} \frac{u}{18} d u

Take the constant out:

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

= \frac{1}{18} \cdot \int_{64}^{73} u d u

Apply the Power Rule

= \frac{1}{18} [\frac{2}{3} u^{\frac{3}{2}}]_{64}^{73}

Compute the boundaries:

\frac{146 73}{3} - \frac{1024}{3}

= \frac{1}{18} (\frac{146 73}{3} - \frac{1024}{3})

Simplify

= \frac{73 73 - 512}{27}

d er i v a t i v e tan (3 t)

\int 2 t 2 t - 3 d t

x^{^{'}} (t) = (\frac{1}{3} x) (1 - \frac{x}{3}), x (0) = 6

\int_{1}^{2} 21 x^{2} 4^{x^{3}} d x

\int 2 x^{2} y + 4 d y

What is the integral from 0 to 1 of sqrt(64t^2+9t^4) ?
The integral from 0 to 1 of sqrt(64t^2+9t^4) is (73sqrt(73)-512)/(27)