What is the integral from 0 to 6 of 2xsqrt(x^2+64) ?

The integral from 0 to 6 of 2xsqrt(x^2+64) is 976/3

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integral from 0 to 6 of 2xsqrt(x^2+64)

\int_{0}^{6} 2 x x^{2} + 64 d x

\frac{976}{3}

Decimal

325.33333 \dots

Solution steps

\int_{0}^{6} 2 x x^{2} + 64 d x

Take the constant out:

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

= 2 \cdot \int_{0}^{6} x x^{2} + 64 d x

Apply u-substitution

: \int_{8}^{10} u^{2} d u

= 2 \cdot \int_{8}^{10} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{8}^{10}

= 2 [\frac{u ^{3}}{3}]_{8}^{10}

Compute the boundaries:

\frac{488}{3}

= 2 \cdot \frac{488}{3}

Simplify

= \frac{976}{3}

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\int_{0}^{1} \frac{x ^{3}}{3} d x

What is the integral from 0 to 6 of 2xsqrt(x^2+64) ?
The integral from 0 to 6 of 2xsqrt(x^2+64) is 976/3