What is the integral from 0 to 1 of xsqrt(1+8x^2) ?

The integral from 0 to 1 of xsqrt(1+8x^2) is 13/12

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integral from 0 to 1 of xsqrt(1+8x^2)

\int_{0}^{1} x 1 + 8 x^{2} d x

\frac{13}{12}

Decimal

1.08333 \dots

Solution steps

\int_{0}^{1} x 1 + 8 x^{2} d x

Apply u-substitution

: \int_{1}^{9} \frac{u}{16} d u

= \int_{1}^{9} \frac{u}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int_{1}^{9} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

= \frac{1}{16} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

Compute the boundaries:

\frac{52}{3}

= \frac{1}{16} \cdot \frac{52}{3}

Simplify

= \frac{13}{12}

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What is the integral from 0 to 1 of xsqrt(1+8x^2) ?
The integral from 0 to 1 of xsqrt(1+8x^2) is 13/12