What is the integral from 0 to 1 of sqrt(9x^4+16x^2) ?

The integral from 0 to 1 of sqrt(9x^4+16x^2) is 61/27

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integral from 0 to 1 of sqrt(9x^4+16x^2)

\int_{0}^{1} 9 x^{4} + 16 x^{2} d x

\frac{61}{27}

Decimal

2.25925 \dots

Solution steps

\int_{0}^{1} 9 x^{4} + 16 x^{2} d x

Factor

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

= \int_{0}^{1} x^{2} 9 x^{2} + 16 d x

x^{2} = (x),

assuming

x \geq 0

= \int_{0}^{1} x 9 x^{2} + 16 d x

Apply u-substitution

: \int_{16}^{25} \frac{u}{18} d u

= \int_{16}^{25} \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}^{25} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{16}^{25}

= \frac{1}{18} [\frac{2}{3} u^{\frac{3}{2}}]_{16}^{25}

Compute the boundaries:

\frac{122}{3}

= \frac{1}{18} \cdot \frac{122}{3}

Simplify

= \frac{61}{27}

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What is the integral from 0 to 1 of sqrt(9x^4+16x^2) ?
The integral from 0 to 1 of sqrt(9x^4+16x^2) is 61/27