What is the integral from 4 to 6 of 30x^3sqrt(1+x^2) ?

The integral from 4 to 6 of 30x^3sqrt(1+x^2) is 7844sqrt(37)-1564sqrt(17)

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integral from 4 to 6 of 30x^3sqrt(1+x^2)

\int_{4}^{6} 30 x^{3} 1 + x^{2} d x

784437 - 156417

Decimal Notation

41264.65208 \dots

Solution steps

\int_{4}^{6} 30 x^{3} 1 + x^{2} d x

Take the constant out:

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

= 30 \cdot \int_{4}^{6} x^{3} 1 + x^{2} d x

Apply u-substitution

: \int_{17}^{37} \frac{( u - 1 ) u}{2} d u

= 30 \cdot \int_{17}^{37} \frac{( u - 1 ) u}{2} d u

Take the constant out:

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

= 30 \cdot \frac{1}{2} \cdot \int_{17}^{37} (u - 1) u d u

Expand

(u - 1) u : u^{\frac{3}{2}} - u

= 30 \cdot \frac{1}{2} \cdot \int_{17}^{37} u^{\frac{3}{2}} - u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 30 \cdot \frac{1}{2} (\int_{17}^{37} u^{\frac{3}{2}} d u - \int_{17}^{37} u d u)

\int_{17}^{37} u^{\frac{3}{2}} d u = \frac{2738 37 - 578 17}{5}

\int_{17}^{37} u d u = \frac{74 37 - 34 17}{3}

= 30 \cdot \frac{1}{2} (\frac{2738 37 - 578 17}{5} - \frac{74 37 - 34 17}{3})

Simplify

30 \cdot \frac{1}{2} (\frac{2738 37 - 578 17}{5} - \frac{74 37 - 34 17}{3}) : 784437 - 156417

= 784437 - 156417

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What is the integral from 4 to 6 of 30x^3sqrt(1+x^2) ?
The integral from 4 to 6 of 30x^3sqrt(1+x^2) is 7844sqrt(37)-1564sqrt(17)