What is the integral from 0 to 1 of xsqrt(63x^2+1) ?

The integral from 0 to 1 of xsqrt(63x^2+1) is 73/27

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integral from 0 to 1 of xsqrt(63x^2+1)

\int_{0}^{1} x 63 x^{2} + 1 d x

\frac{73}{27}

Decimal

2.70370 \dots

Solution steps

\int_{0}^{1} x 63 x^{2} + 1 d x

Apply u-substitution

= \int_{1}^{64} \frac{u}{126} d u

Take the constant out:

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

= \frac{1}{126} \cdot \int_{1}^{64} u d u

Apply the Power Rule

= \frac{1}{126} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{64}

Compute the boundaries:

\frac{1022}{3}

= \frac{1}{126} \cdot \frac{1022}{3}

Simplify

= \frac{73}{27}

What is the integral from 0 to 1 of xsqrt(63x^2+1) ?
The integral from 0 to 1 of xsqrt(63x^2+1) is 73/27