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

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

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

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

\frac{13 13 - 8}{27}

Decimal

1.43970 \dots

Solution steps

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

Factor

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

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

x^{2} = (x),

assuming

x \geq 0

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

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_{0}^{5} sin (\frac{nπ x}{5}) d x

\int_{0}^{4} 3 x^{2} - x + 1 d x

\int_{- 1}^{2} 3 x^{2} + 1 d x

\int_{0}^{2 π} z cos (n z) d z

\int_{0}^{\frac{2 π}{3}} e^{2 i t} d t

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