What is the integral from 0 to 1 of 11xsqrt(x^2+36) ?

The integral from 0 to 1 of 11xsqrt(x^2+36) is 11((37sqrt(37))/3-72)

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integral from 0 to 1 of 11xsqrt(x^2+36)

\int_{0}^{1} 11 x x^{2} + 36 d x

11 (\frac{37 37}{3} - 72)

Decimal Notation

33.22811 \dots

Solution steps

\int_{0}^{1} 11 x x^{2} + 36 d x

Take the constant out:

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

= 11 \cdot \int_{0}^{1} x x^{2} + 36 d x

Apply u-substitution

: \int_{6}^{37} u^{2} d u

= 11 \cdot \int_{6}^{37} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{6}^{37}

= 11 [\frac{u ^{3}}{3}]_{6}^{37}

Compute the boundaries:

\frac{37 37}{3} - 72

= 11 (\frac{37 37}{3} - 72)

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What is the integral from 0 to 1 of 11xsqrt(x^2+36) ?
The integral from 0 to 1 of 11xsqrt(x^2+36) is 11((37sqrt(37))/3-72)