What is the integral from 0 to 1 of xsqrt(35x^2+1) ?

The integral from 0 to 1 of xsqrt(35x^2+1) is 43/21

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integral from 0 to 1 of xsqrt(35x^2+1)

\int_{0}^{1} x 35 x^{2} + 1 d x

\frac{43}{21}

Decimal

2.04761 \dots

Solution steps

\int_{0}^{1} x 35 x^{2} + 1 d x

Apply u-substitution

: \int_{1}^{36} \frac{u}{70} d u

= \int_{1}^{36} \frac{u}{70} d u

Take the constant out:

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

= \frac{1}{70} \cdot \int_{1}^{36} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{36}

= \frac{1}{70} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{36}

Compute the boundaries:

\frac{430}{3}

= \frac{1}{70} \cdot \frac{430}{3}

Simplify

= \frac{43}{21}

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What is the integral from 0 to 1 of xsqrt(35x^2+1) ?
The integral from 0 to 1 of xsqrt(35x^2+1) is 43/21