What is the integral from 0 to 2 of 6x^3sqrt(x^4+9) ?

The integral from 0 to 2 of 6x^3sqrt(x^4+9) is 98

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integral from 0 to 2 of 6x^3sqrt(x^4+9)

\int_{0}^{2} 6 x^{3} x^{4} + 9 d x

98

Solution steps

\int_{0}^{2} 6 x^{3} x^{4} + 9 d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int_{0}^{2} x^{3} x^{4} + 9 d x

Apply u-substitution

: \int_{9}^{25} \frac{u}{4} d u

= 6 \cdot \int_{9}^{25} \frac{u}{4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \frac{1}{4} \cdot \int_{9}^{25} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{25}

= 6 \cdot \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{25}

Simplify

6 \cdot \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{25} : \frac{3}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{25}

= \frac{3}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{25}

Compute the boundaries:

\frac{196}{3}

= \frac{3}{2} \cdot \frac{196}{3}

Simplify

= 98

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What is the integral from 0 to 2 of 6x^3sqrt(x^4+9) ?
The integral from 0 to 2 of 6x^3sqrt(x^4+9) is 98