What is the integral from 0 to 5 of sqrt(9t^4+16t^2) ?

The integral from 0 to 5 of sqrt(9t^4+16t^2) is (241sqrt(241)-64)/(27)

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integral from 0 to 5 of sqrt(9t^4+16t^2)

\int_{0}^{5} 9 t^{4} + 16 t^{2} d t

\frac{241 241 - 64}{27}

Decimal

136.19726 \dots

Solution steps

\int_{0}^{5} 9 t^{4} + 16 t^{2} d t

Factor

9 t^{4} + 16 t^{2} : t^{2} 9 t^{2} + 16

= \int_{0}^{5} t^{2} 9 t^{2} + 16 d t

t^{2} = (t),

assuming

t \geq 0

= \int_{0}^{5} t 9 t^{2} + 16 d t

Apply u-substitution

: \int_{16}^{241} \frac{u}{18} d u

= \int_{16}^{241} \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_{16}^{241} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{16}^{241}

= \frac{1}{18} [\frac{2}{3} u^{\frac{3}{2}}]_{16}^{241}

Compute the boundaries:

\frac{482 241}{3} - \frac{128}{3}

= \frac{1}{18} (\frac{482 241}{3} - \frac{128}{3})

Simplify

= \frac{241 241 - 64}{27}

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What is the integral from 0 to 5 of sqrt(9t^4+16t^2) ?
The integral from 0 to 5 of sqrt(9t^4+16t^2) is (241sqrt(241)-64)/(27)