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

The integral from 0 to 1 of sqrt(9t^4+4t^2) is (13sqrt(13)-8)/(27)

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

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

\frac{13 13 - 8}{27}

Decimal

1.43970 \dots

Solution steps

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

Factor

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

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

t^{2} = (t),

assuming

t \geq 0

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

Apply u-substitution

: \int_{4}^{13} \frac{u}{18} d u

= \int_{4}^{13} \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}^{13} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{13}

= \frac{1}{18} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{13}

Compute the boundaries:

\frac{26 13}{3} - \frac{16}{3}

= \frac{1}{18} (\frac{26 13}{3} - \frac{16}{3})

Simplify

= \frac{13 13 - 8}{27}

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

\frac{d}{d x} (- x^{- 2})

\int e^{x} - 1 d x

\frac{d}{d x} (x^{4} - 8 x^{2})

\int_{1}^{e^{3}} \frac{ln ^{4} ( x ^{2} )}{x} d x

What is the integral from 0 to 1 of sqrt(9t^4+4t^2) ?
The integral from 0 to 1 of sqrt(9t^4+4t^2) is (13sqrt(13)-8)/(27)