What is the integral from 0 to 1 of xsqrt(27x^2+9) ?

The integral from 0 to 1 of xsqrt(27x^2+9) is 7/3

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integral from 0 to 1 of xsqrt(27x^2+9)

\int_{0}^{1} x 27 x^{2} + 9 d x

\frac{7}{3}

Decimal

2.33333 \dots

Solution steps

\int_{0}^{1} x 27 x^{2} + 9 d x

Apply u-substitution

: \int_{9}^{36} \frac{u}{54} d u

= \int_{9}^{36} \frac{u}{54} d u

Take the constant out:

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

= \frac{1}{54} \cdot \int_{9}^{36} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{36}

= \frac{1}{54} [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{36}

Compute the boundaries:

126

= \frac{1}{54} \cdot 126

Simplify

= \frac{7}{3}

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What is the integral from 0 to 1 of xsqrt(27x^2+9) ?
The integral from 0 to 1 of xsqrt(27x^2+9) is 7/3