What is the integral from 1 to 3 of 2xsqrt(x^2+9) ?

The integral from 1 to 3 of 2xsqrt(x^2+9) is 2(18sqrt(2)-(10sqrt(10))/3)

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integral from 1 to 3 of 2xsqrt(x^2+9)

\int_{1}^{3} 2 x x^{2} + 9 d x

2 (182 - \frac{10 10}{3})

Decimal Notation

29.82983 \dots

Solution steps

\int_{1}^{3} 2 x x^{2} + 9 d x

Take the constant out:

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

= 2 \cdot \int_{1}^{3} x x^{2} + 9 d x

Apply u-substitution

: \int_{10}^{32} u^{2} d u

= 2 \cdot \int_{10}^{32} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{10}^{32}

= 2 [\frac{u ^{3}}{3}]_{10}^{32}

Compute the boundaries:

182 - \frac{10 10}{3}

= 2 (182 - \frac{10 10}{3})

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What is the integral from 1 to 3 of 2xsqrt(x^2+9) ?
The integral from 1 to 3 of 2xsqrt(x^2+9) is 2(18sqrt(2)-(10sqrt(10))/3)