What is the integral from 0 to 1 of xsqrt(39x^2+25) ?

The integral from 0 to 1 of xsqrt(39x^2+25) is 43/13

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integral from 0 to 1 of xsqrt(39x^2+25)

\int_{0}^{1} x 39 x^{2} + 25 d x

\frac{43}{13}

Decimal

3.30769 \dots

Solution steps

\int_{0}^{1} x 39 x^{2} + 25 d x

Apply u-substitution

= \int_{25}^{64} \frac{u}{78} d u

Take the constant out:

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

= \frac{1}{78} \cdot \int_{25}^{64} u d u

Apply the Power Rule

= \frac{1}{78} [\frac{2}{3} u^{\frac{3}{2}}]_{25}^{64}

Compute the boundaries:

258

= \frac{1}{78} \cdot 258

Simplify

= \frac{43}{13}

What is the integral from 0 to 1 of xsqrt(39x^2+25) ?
The integral from 0 to 1 of xsqrt(39x^2+25) is 43/13