What is the integral from 1 to 3 of sqrt(x^2+x^4) ?

The integral from 1 to 3 of sqrt(x^2+x^4) is (10sqrt(10)-2sqrt(2))/3

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integral from 1 to 3 of sqrt(x^2+x^4)

\int_{1}^{3} x^{2} + x^{4} d x

\frac{10 10 - 2 2}{3}

Decimal

9.59811 \dots

Solution steps

\int_{1}^{3} x^{2} + x^{4} d x

Simplify

x^{2} + x^{4} : x^{2} 1 + x^{2}

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

x^{2} = (x),

assuming

x \geq 0

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

Apply u-substitution

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

Apply the Power Rule

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

Compute the boundaries:

\frac{10 10 - 2 2}{3}

= \frac{10 10 - 2 2}{3}

What is the integral from 1 to 3 of sqrt(x^2+x^4) ?
The integral from 1 to 3 of sqrt(x^2+x^4) is (10sqrt(10)-2sqrt(2))/3