What is the integral from 1 to 2 of 21x^24^{x^3} ?

The integral from 1 to 2 of 21x^24^{x^3} is (229362)/(ln(2))

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integral from 1 to 2 of 21x^24^{x^3}

\int_{1}^{2} 21 x^{2} 4^{x^{3}} d x

\frac{229362}{ln ( 2 )}

Decimal Notation

330899.41996 \dots

Solution steps

\int_{1}^{2} 21 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

= 21 \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

= 21 \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

= 21 \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}

= 21 \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

= 21 \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

21 \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{7}{ln ( 2 )} [\frac{u ^{2}}{2 ^{\frac{l n ( 2 u ) + l n ( 2 )}{l n ( 2 )}}} + \frac{1}{4} u]_{4}^{65536}

= \frac{7}{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{7}{ln ( 2 )} \cdot 32766

Simplify

= \frac{229362}{ln ( 2 )}

\int 2 x^{2} y + 4 d y

\frac{d}{d x} (4 x^{2} + 2)

\int_{0}^{16} 2 x - \frac{x}{2} d x

x \to - 1 lim (∣ x ∣)

y^{^{''}} - 4 y^{^{'}} + 13 y = 0

What is the integral from 1 to 2 of 21x^24^{x^3} ?
The integral from 1 to 2 of 21x^24^{x^3} is (229362)/(ln(2))