What is the integral from 0 to 1 of 7xsqrt(x^2+16) ?

The integral from 0 to 1 of 7xsqrt(x^2+16) is (7(17sqrt(17)-64))/3

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integral from 0 to 1 of 7xsqrt(x^2+16)

\int_{0}^{1} 7 x x^{2} + 16 d x

\frac{7 ( 17 17 - 64 )}{3}

Decimal Notation

14.21652 \dots

Solution steps

\int_{0}^{1} 7 x x^{2} + 16 d x

Take the constant out:

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

= 7 \cdot \int_{0}^{1} x x^{2} + 16 d x

Apply u-substitution

: \int_{4}^{17} u^{2} d u

= 7 \cdot \int_{4}^{17} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{4}^{17}

= 7 [\frac{u ^{3}}{3}]_{4}^{17}

Compute the boundaries:

\frac{17 17}{3} - \frac{64}{3}

= 7 (\frac{17 17}{3} - \frac{64}{3})

Simplify

= \frac{7 ( 17 17 - 64 )}{3}

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What is the integral from 0 to 1 of 7xsqrt(x^2+16) ?
The integral from 0 to 1 of 7xsqrt(x^2+16) is (7(17sqrt(17)-64))/3