What is the integral from-1 to 2 of 4^xe^{2x} ?

The integral from-1 to 2 of 4^xe^{2x} is

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integral from-1 to 2 of 4^xe^{2x}

\int_{- 1}^{2} 4^{x} e^{2 x} d x

\frac{2 ^{\frac{6 l n ( 2 ) + 6}{l n ( 2 )}} ln ( 2 ) - ln ( 2 )}{4 \cdot 2 ^{\frac{l n ( 2 ) + 1}{l n ( 2 )}} \cdot 2 ^{\frac{1}{l n ( 2 )}} ln ^{2} ( 2 ) + 4 \cdot 2 ^{\frac{l n ( 2 ) + 1}{l n ( 2 )}} \cdot 2 ^{\frac{1}{l n ( 2 )}} ln ( 2 )}

Decimal Notation

257.96238 \dots

Solution steps

\int_{- 1}^{2} 4^{x} e^{2 x} d x

Apply u-substitution

: \int_{\frac{1}{4}}^{16} \frac{u u ^{\frac{1}{l n ( 2 )}}}{ln ( 2 ) \cdot 2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}}} d u

= \int_{\frac{1}{4}}^{16} \frac{u u ^{\frac{1}{l n ( 2 )}}}{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}{ln ( 2 )} \cdot \int_{\frac{1}{4}}^{16} \frac{u u ^{\frac{1}{l n ( 2 )}}}{2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}}} d u

Apply u-substitution

: \int_{\frac{1}{2}}^{32} \frac{v ^{\frac{1}{l n ( 2 )}}}{4 \cdot 2 ^{\frac{1}{l n ( 2 )}}} d v

= \frac{1}{ln ( 2 )} \cdot \int_{\frac{1}{2}}^{32} \frac{v ^{\frac{1}{l n ( 2 )}}}{4 \cdot 2 ^{\frac{1}{l n ( 2 )}}} d v

Take the constant out:

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

= \frac{1}{ln ( 2 )} \cdot \frac{1}{4 \cdot 2 ^{\frac{1}{l n ( 2 )}}} \cdot \int_{\frac{1}{2}}^{32} v^{\frac{1}{l n ( 2 )}} d v

Apply the Power Rule

: [\frac{1}{1 + ln ( 2 )} ln (2) v^{\frac{1 + l n ( 2 )}{l n ( 2 )}}]_{\frac{1}{2}}^{32}

= \frac{1}{ln ( 2 )} \cdot \frac{1}{4 \cdot 2 ^{\frac{1}{l n ( 2 )}}} [\frac{1}{1 + ln ( 2 )} ln (2) v^{\frac{1 + l n ( 2 )}{l n ( 2 )}}]_{\frac{1}{2}}^{32}

Simplify

\frac{1}{ln ( 2 )} \cdot \frac{1}{4 \cdot 2 ^{\frac{1}{l n ( 2 )}}} [\frac{1}{1 + ln ( 2 )} ln (2) v^{\frac{1 + l n ( 2 )}{l n ( 2 )}}]_{\frac{1}{2}}^{32} : \frac{1}{4 \cdot 2 ^{\frac{1}{l n ( 2 )}} ln ( 2 )} [\frac{1}{1 + ln ( 2 )} ln (2) v^{\frac{1 + l n ( 2 )}{l n ( 2 )}}]_{\frac{1}{2}}^{32}

= \frac{1}{4 \cdot 2 ^{\frac{1}{l n ( 2 )}} ln ( 2 )} [\frac{1}{1 + ln ( 2 )} ln (2) v^{\frac{1 + l n ( 2 )}{l n ( 2 )}}]_{\frac{1}{2}}^{32}

Compute the boundaries:

\frac{2 ^{\frac{6 l n ( 2 ) + 6}{l n ( 2 )}} ln ( 2 ) - ln ( 2 )}{2 ^{\frac{l n ( 2 ) + 1}{l n ( 2 )}} ( ln ( 2 ) + 1 )}

= \frac{1}{4 \cdot 2 ^{\frac{1}{l n ( 2 )}} ln ( 2 )} \cdot \frac{2 ^{\frac{6 l n ( 2 ) + 6}{l n ( 2 )}} ln ( 2 ) - ln ( 2 )}{2 ^{\frac{l n ( 2 ) + 1}{l n ( 2 )}} ( ln ( 2 ) + 1 )}

Simplify

= \frac{2 ^{\frac{6 l n ( 2 ) + 6}{l n ( 2 )}} ln ( 2 ) - ln ( 2 )}{4 \cdot 2 ^{\frac{l n ( 2 ) + 1}{l n ( 2 )}} \cdot 2 ^{\frac{1}{l n ( 2 )}} ln ^{2} ( 2 ) + 4 \cdot 2 ^{\frac{l n ( 2 ) + 1}{l n ( 2 )}} \cdot 2 ^{\frac{1}{l n ( 2 )}} ln ( 2 )}

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What is the integral from-1 to 2 of 4^xe^{2x} ?
The integral from-1 to 2 of 4^xe^{2x} is