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

The integral from 0 to 2 of sqrt(4t^2+9t^4) is (8(10sqrt(10)-1))/(27)

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

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

\frac{8 ( 10 10 - 1 )}{27}

Decimal Notation

9.07341 \dots

Solution steps

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

Factor

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

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

t^{2} = (t),

assuming

t \geq 0

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

Apply u-substitution

: \int_{4}^{40} \frac{u}{18} d u

= \int_{4}^{40} \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_{4}^{40} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{40}

= \frac{1}{18} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{40}

Compute the boundaries:

\frac{160 10}{3} - \frac{16}{3}

= \frac{1}{18} (\frac{160 10}{3} - \frac{16}{3})

Simplify

= \frac{8 ( 10 10 - 1 )}{27}

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What is the integral from 0 to 2 of sqrt(4t^2+9t^4) ?
The integral from 0 to 2 of sqrt(4t^2+9t^4) is (8(10sqrt(10)-1))/(27)