What is the integral from 1 to 2 of 12x^23^{x^3} ?

The integral from 1 to 2 of 12x^23^{x^3} is (26232)/(ln(3))

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integral from 1 to 2 of 12x^23^{x^3}

\int_{1}^{2} 12 x^{2} 3^{x^{3}} d x

\frac{26232}{ln ( 3 )}

Decimal Notation

23877.39539 \dots

Solution steps

\int_{1}^{2} 12 x^{2} \cdot 3^{x^{3}} d x

Take the constant out:

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

= 12 \cdot \int_{1}^{2} x^{2} \cdot 3^{x^{3}} d x

Apply u-substitution

: \int_{3}^{6561} \frac{u}{ln ( 3 ) \cdot 3 ^{\frac{l n ( u ) + l n ( 3 )}{l n ( 3 )}}} d u

= 12 \cdot \int_{3}^{6561} \frac{u}{ln ( 3 ) \cdot 3 ^{\frac{l n ( u ) + l n ( 3 )}{l n ( 3 )}}} d u

Take the constant out:

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

= 12 \cdot \frac{1}{ln ( 3 )} \cdot \int_{3}^{6561} \frac{u}{3 ^{\frac{l n ( u ) + l n ( 3 )}{l n ( 3 )}}} d u

Apply Integration By Parts

: [\frac{u ^{2}}{2 \cdot 3 ^{\frac{l n ( u ) + l n ( 3 )}{l n ( 3 )}}} - \int - \frac{1}{6} d u]_{3}^{6561}

= 12 \cdot \frac{1}{ln ( 3 )} [\frac{u ^{2}}{2 \cdot 3 ^{\frac{l n ( u ) + l n ( 3 )}{l n ( 3 )}}} - \int - \frac{1}{6} d u]_{3}^{6561}

\int - \frac{1}{6} d u = - \frac{1}{6} u

= 12 \cdot \frac{1}{ln ( 3 )} [\frac{u ^{2}}{2 \cdot 3 ^{\frac{l n ( u ) + l n ( 3 )}{l n ( 3 )}}} - (- \frac{1}{6} u)]_{3}^{6561}

Simplify

12 \cdot \frac{1}{ln ( 3 )} [\frac{u ^{2}}{2 \cdot 3 ^{\frac{l n ( u ) + l n ( 3 )}{l n ( 3 )}}} - (- \frac{1}{6} u)]_{3}^{6561} : \frac{12}{ln ( 3 )} [\frac{u ^{2}}{2 \cdot 3 ^{\frac{l n ( u ) + l n ( 3 )}{l n ( 3 )}}} + \frac{1}{6} u]_{3}^{6561}

= \frac{12}{ln ( 3 )} [\frac{u ^{2}}{2 \cdot 3 ^{\frac{l n ( u ) + l n ( 3 )}{l n ( 3 )}}} + \frac{1}{6} u]_{3}^{6561}

Compute the boundaries:

2186

= \frac{12}{ln ( 3 )} \cdot 2186

Simplify

= \frac{26232}{ln ( 3 )}

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\int_{- 1}^{0} y + 1 d y

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\int \frac{1}{486} (81 - x^{2}) d x

n = 1 \sum \infty \frac{1}{n ^{2} + 4}

What is the integral from 1 to 2 of 12x^23^{x^3} ?
The integral from 1 to 2 of 12x^23^{x^3} is (26232)/(ln(3))