What is the integral from 0 to 1 of xsqrt(99x^2+1) ?

The integral from 0 to 1 of xsqrt(99x^2+1) is 37/11

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integral from 0 to 1 of xsqrt(99x^2+1)

\int_{0}^{1} x 99 x^{2} + 1 d x

\frac{37}{11}

Decimal

3.36363 \dots

Solution steps

\int_{0}^{1} x 99 x^{2} + 1 d x

Apply u-substitution

: \int_{1}^{100} \frac{u}{198} d u

= \int_{1}^{100} \frac{u}{198} d u

Take the constant out:

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

= \frac{1}{198} \cdot \int_{1}^{100} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{100}

= \frac{1}{198} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{100}

Compute the boundaries:

666

= \frac{1}{198} \cdot 666

Simplify

= \frac{37}{11}

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What is the integral from 0 to 1 of xsqrt(99x^2+1) ?
The integral from 0 to 1 of xsqrt(99x^2+1) is 37/11