What is the integral from 1 to 2 of 18x^24^{x^3} ?

The integral from 1 to 2 of 18x^24^{x^3} is (196596)/(ln(2))

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integral from 1 to 2 of 18x^24^{x^3}

\int_{1}^{2} 18 x^{2} 4^{x^{3}} d x

\frac{196596}{ln ( 2 )}

Decimal

283628.07425 \dots

Solution steps

\int_{1}^{2} 18 x^{2} \cdot 4^{x^{3}} d x

Take the constant out:

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

= 18 \cdot \int_{1}^{2} x^{2} \cdot 4^{x^{3}} d x

Apply u-substitution

: \int_{4}^{65536} \frac{u}{3 ln ( 2 ) \cdot 2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}}} d u

= 18 \cdot \int_{4}^{65536} \frac{u}{3 ln ( 2 ) \cdot 2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}}} d u

Take the constant out:

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

= 18 \cdot \frac{1}{3 ln ( 2 )} \cdot \int_{4}^{65536} \frac{u}{2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}}} d u

Apply Integration By Parts

: [\frac{u ^{2}}{2 ^{\frac{l n ( 2 u ) + l n ( 2 )}{l n ( 2 )}}} - \int - \frac{1}{4} d u]_{4}^{65536}

= 18 \cdot \frac{1}{3 ln ( 2 )} [\frac{u ^{2}}{2 ^{\frac{l n ( 2 u ) + l n ( 2 )}{l n ( 2 )}}} - \int - \frac{1}{4} d u]_{4}^{65536}

\int - \frac{1}{4} d u = - \frac{1}{4} u

= 18 \cdot \frac{1}{3 ln ( 2 )} [\frac{u ^{2}}{2 ^{\frac{l n ( 2 u ) + l n ( 2 )}{l n ( 2 )}}} - (- \frac{1}{4} u)]_{4}^{65536}

Simplify

18 \cdot \frac{1}{3 ln ( 2 )} [\frac{u ^{2}}{2 ^{\frac{l n ( 2 u ) + l n ( 2 )}{l n ( 2 )}}} - (- \frac{1}{4} u)]_{4}^{65536} : \frac{6}{ln ( 2 )} [\frac{u ^{2}}{2 ^{\frac{l n ( 2 u ) + l n ( 2 )}{l n ( 2 )}}} + \frac{1}{4} u]_{4}^{65536}

= \frac{6}{ln ( 2 )} [\frac{u ^{2}}{2 ^{\frac{l n ( 2 u ) + l n ( 2 )}{l n ( 2 )}}} + \frac{1}{4} u]_{4}^{65536}

Compute the boundaries:

32766

= \frac{6}{ln ( 2 )} \cdot 32766

Simplify

= \frac{196596}{ln ( 2 )}

\int \frac{x ^{2}}{x + 1} d x

x \to 0 - lim (1 + e^{\frac{1}{x}})

3 t^{2} + 15 t + 15

y = (5 x^{2} - 1)^{- 3}

\int_{- \infty}^{0} x e^{8 x} d x

What is the integral from 1 to 2 of 18x^24^{x^3} ?
The integral from 1 to 2 of 18x^24^{x^3} is (196596)/(ln(2))