What is the integral from 0 to 1 of xsqrt(5x^2+4) ?

The integral from 0 to 1 of xsqrt(5x^2+4) is 19/15

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integral from 0 to 1 of xsqrt(5x^2+4)

\int_{0}^{1} x 5 x^{2} + 4 d x

\frac{19}{15}

Decimal

1.26666 \dots

Solution steps

\int_{0}^{1} x 5 x^{2} + 4 d x

Apply u-substitution

: \int_{4}^{9} \frac{u}{10} d u

= \int_{4}^{9} \frac{u}{10} d u

Take the constant out:

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

= \frac{1}{10} \cdot \int_{4}^{9} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{9}

= \frac{1}{10} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{9}

Compute the boundaries:

\frac{38}{3}

= \frac{1}{10} \cdot \frac{38}{3}

Simplify

= \frac{19}{15}

\int_{0}^{π} 3 cos (x) d x

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

\int_{1}^{3} 3 - x d x

\int_{1}^{3} \frac{3}{x ( x + 1 )} d x

\int_{0}^{1} 3 y^{2} d y

What is the integral from 0 to 1 of xsqrt(5x^2+4) ?
The integral from 0 to 1 of xsqrt(5x^2+4) is 19/15