What is the integral from 0 to 8 of 2xsqrt(x^2+36) ?

The integral from 0 to 8 of 2xsqrt(x^2+36) is 1568/3

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integral from 0 to 8 of 2xsqrt(x^2+36)

\int_{0}^{8} 2 x x^{2} + 36 d x

\frac{1568}{3}

Solution steps

\int_{0}^{8} 2 x x^{2} + 36 d x

Take the constant out:

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

= 2 \cdot \int_{0}^{8} x x^{2} + 36 d x

Apply u-substitution

: \int_{6}^{10} u^{2} d u

= 2 \cdot \int_{6}^{10} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{6}^{10}

= 2 [\frac{u ^{3}}{3}]_{6}^{10}

Compute the boundaries:

\frac{784}{3}

= 2 \cdot \frac{784}{3}

Simplify

= \frac{1568}{3}

\int_{0}^{y^{2}} sin (\frac{x}{y}) d x

\int_{0}^{5} 9 e^{3 x} d x

\int_{0}^{4} \frac{p}{9 + p ^{2}} d p

\int_{0}^{4} 2 x^{2} + 5 d x

What is the integral from 0 to 8 of 2xsqrt(x^2+36) ?
The integral from 0 to 8 of 2xsqrt(x^2+36) is 1568/3