What is the integral from 1 to 2 of 24x^22^{x^3} ?

The integral from 1 to 2 of 24x^22^{x^3} is (2032)/(ln(2))

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integral from 1 to 2 of 24x^22^{x^3}

\int_{1}^{2} 24 x^{2} 2^{x^{3}} d x

\frac{2032}{ln ( 2 )}

Decimal Notation

2931.55632 \dots

Solution steps

\int_{1}^{2} 24 x^{2} \cdot 2^{x^{3}} d x

Take the constant out:

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

= 24 \cdot \int_{1}^{2} x^{2} \cdot 2^{x^{3}} d x

Apply u-substitution

: \int_{1}^{8} \frac{2 ^{u}}{3} d u

= 24 \cdot \int_{1}^{8} \frac{2 ^{u}}{3} d u

Take the constant out:

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

= 24 \cdot \frac{1}{3} \cdot \int_{1}^{8} 2^{u} d u

\int a^{x} d x = \frac{a ^{x}}{ln a}

= 24 \cdot \frac{1}{3} [\frac{2 ^{u}}{ln ( 2 )}]_{1}^{8}

Simplify

24 \cdot \frac{1}{3} [\frac{2 ^{u}}{ln ( 2 )}]_{1}^{8} : 8 [\frac{2 ^{u}}{ln ( 2 )}]_{1}^{8}

= 8 [\frac{2 ^{u}}{ln ( 2 )}]_{1}^{8}

Compute the boundaries:

\frac{254}{ln ( 2 )}

= 8 \cdot \frac{254}{ln ( 2 )}

Simplify

= \frac{2032}{ln ( 2 )}

\int 6^{x - 2} d x

\frac{d}{d x} (12 cos (2 x))

\frac{3}{s} + \frac{8}{s ( s + 1 )}

\frac{d}{d x} (7 x + 5)

(x^{0.5} + 4) (x^{2} + x)

What is the integral from 1 to 2 of 24x^22^{x^3} ?
The integral from 1 to 2 of 24x^22^{x^3} is (2032)/(ln(2))