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

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

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

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

\frac{13 13 - 8}{27}

Decimal Notation

1.43970 \dots

Solution steps

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

Factor

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

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

t^{2} = (t),

assuming

t \geq 0

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

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

\frac{x ^{2} + 2 x + 1}{x ^{2} - 2 x + 1}

- \frac{7}{1 + x ^{2}}

n = 1 \sum \infty \frac{n}{2 n ^{3} + 3}

\frac{4 x + 5}{4 x - 5}

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