What is the integral from 1 to 2 of 7x^2sqrt(8+x^3) ?

The integral from 1 to 2 of 7x^2sqrt(8+x^3) is 518/9

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integral from 1 to 2 of 7x^2sqrt(8+x^3)

\int_{1}^{2} 7 x^{2} 8 + x^{3} d x

\frac{518}{9}

Decimal

57.55555 \dots

Solution steps

\int_{1}^{2} 7 x^{2} 8 + x^{3} d x

Take the constant out:

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

= 7 \cdot \int_{1}^{2} x^{2} 8 + x^{3} d x

Apply u-substitution

: \int_{9}^{16} \frac{u}{3} d u

= 7 \cdot \int_{9}^{16} \frac{u}{3} d u

Take the constant out:

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

= 7 \cdot \frac{1}{3} \cdot \int_{9}^{16} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{16}

= 7 \cdot \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{16}

Simplify

7 \cdot \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{16} : \frac{7}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{16}

= \frac{7}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{16}

Compute the boundaries:

\frac{74}{3}

= \frac{7}{3} \cdot \frac{74}{3}

Simplify

= \frac{518}{9}

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What is the integral from 1 to 2 of 7x^2sqrt(8+x^3) ?
The integral from 1 to 2 of 7x^2sqrt(8+x^3) is 518/9