What is the integral from 0 to 1 of xsqrt(3x^2+1) ?

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

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integral from 0 to 1 of xsqrt(3x^2+1)

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

\frac{7}{9}

Decimal

0.77777 \dots

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Apply the Power Rule

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

Compute the boundaries:

\frac{14}{3}

= \frac{1}{6} \cdot \frac{14}{3}

Simplify

= \frac{7}{9}

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