What is the integral from 1 to 5 of 2x(3x^2+5) ?

The integral from 1 to 5 of 2x(3x^2+5) is 1056

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integral from 1 to 5 of 2x(3x^2+5)

\int_{1}^{5} 2 x (3 x^{2} + 5) d x

1056

Solution steps

\int_{1}^{5} 2 x (3 x^{2} + 5) d x

Take the constant out:

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

= 2 \cdot \int_{1}^{5} x (3 x^{2} + 5) d x

Apply u-substitution

: \int_{8}^{80} \frac{u}{6} d u

= 2 \cdot \int_{8}^{80} \frac{u}{6} d u

Take the constant out:

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

= 2 \cdot \frac{1}{6} \cdot \int_{8}^{80} u d u

Apply the Power Rule

: [\frac{u ^{2}}{2}]_{8}^{80}

= 2 \cdot \frac{1}{6} [\frac{u ^{2}}{2}]_{8}^{80}

Simplify

2 \cdot \frac{1}{6} [\frac{u ^{2}}{2}]_{8}^{80} : \frac{1}{3} [\frac{u ^{2}}{2}]_{8}^{80}

= \frac{1}{3} [\frac{u ^{2}}{2}]_{8}^{80}

Compute the boundaries:

3168

= \frac{1}{3} \cdot 3168

Simplify

= 1056

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What is the integral from 1 to 5 of 2x(3x^2+5) ?
The integral from 1 to 5 of 2x(3x^2+5) is 1056