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

The integral from 1 to 2 of 3*x^2*(x^3+3)^6 is 2781541

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

\int_{1}^{2} 3 \cdot x^{2} \cdot (x^{3} + 3)^{6} d x

2781541

Solution steps

\int_{1}^{2} 3 x^{2} (x^{3} + 3)^{6} d x

Take the constant out:

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

= 3 \cdot \int_{1}^{2} x^{2} (x^{3} + 3)^{6} d x

Apply u-substitution

: \int_{1}^{8} \frac{( u + 3 ) ^{6}}{3} d u

= 3 \cdot \int_{1}^{8} \frac{( u + 3 ) ^{6}}{3} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int_{1}^{8} (u + 3)^{6} d u

Apply u-substitution

: \int_{4}^{11} v^{6} d v

= 3 \cdot \frac{1}{3} \cdot \int_{4}^{11} v^{6} d v

Apply the Power Rule

: [\frac{v ^{7}}{7}]_{4}^{11}

= 3 \cdot \frac{1}{3} [\frac{v ^{7}}{7}]_{4}^{11}

Simplify

3 \cdot \frac{1}{3} [\frac{v ^{7}}{7}]_{4}^{11} : [\frac{v ^{7}}{7}]_{4}^{11}

= [\frac{v ^{7}}{7}]_{4}^{11}

Compute the boundaries:

2781541

= 2781541

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What is the integral from 1 to 2 of 3*x^2*(x^3+3)^6 ?
The integral from 1 to 2 of 3*x^2*(x^3+3)^6 is 2781541