What is the integral from 1 to 2 of x8^{x^2} ?

The integral from 1 to 2 of x8^{x^2} is (2044)/(3ln(2))

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integral from 1 to 2 of x8^{x^2}

\int_{1}^{2} x 8^{x^{2}} d x

\frac{2044}{3 ln ( 2 )}

Decimal

982.95622 \dots

Solution steps

\int_{1}^{2} x \cdot 8^{x^{2}} d x

Apply u-substitution

: \int_{8}^{4096} \frac{u}{3 ln ( 2 ) \cdot 2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}}} d u

= \int_{8}^{4096} \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

= \frac{1}{3 ln ( 2 )} \cdot \int_{8}^{4096} \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]_{8}^{4096}

= \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]_{8}^{4096}

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

= \frac{1}{3 ln ( 2 )} [\frac{u ^{2}}{2 ^{\frac{l n ( 2 u ) + l n ( 2 )}{l n ( 2 )}}} - (- \frac{1}{4} u)]_{8}^{4096}

Simplify

= \frac{1}{3 ln ( 2 )} [\frac{u ^{2}}{2 ^{\frac{l n ( 2 u ) + l n ( 2 )}{l n ( 2 )}}} + \frac{1}{4} u]_{8}^{4096}

Compute the boundaries:

2044

= \frac{1}{3 ln ( 2 )} \cdot 2044

Simplify

= \frac{2044}{3 ln ( 2 )}

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What is the integral from 1 to 2 of x8^{x^2} ?
The integral from 1 to 2 of x8^{x^2} is (2044)/(3ln(2))