What is the integral from 1 to 2 of 12x^22^{x^3} ?

The integral from 1 to 2 of 12x^22^{x^3} is (1016)/(ln(2))

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integral from 1 to 2 of 12x^22^{x^3}

\int_{1}^{2} 12 x^{2} 2^{x^{3}} d x

\frac{1016}{ln ( 2 )}

Decimal Notation

1465.77816 \dots

Solution steps

\int_{1}^{2} 12 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

= 12 \cdot \int_{1}^{2} x^{2} \cdot 2^{x^{3}} d x

Apply u-substitution

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

= 12 \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

= 12 \cdot \frac{1}{3} \cdot \int_{1}^{8} 2^{u} d u

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

= 12 \cdot \frac{1}{3} [\frac{2 ^{u}}{ln ( 2 )}]_{1}^{8}

Simplify

12 \cdot \frac{1}{3} [\frac{2 ^{u}}{ln ( 2 )}]_{1}^{8} : 4 [\frac{2 ^{u}}{ln ( 2 )}]_{1}^{8}

= 4 [\frac{2 ^{u}}{ln ( 2 )}]_{1}^{8}

Compute the boundaries:

\frac{254}{ln ( 2 )}

= 4 \cdot \frac{254}{ln ( 2 )}

Simplify

= \frac{1016}{ln ( 2 )}

y^{^{''}} + 26 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = 0

f (x) = (x + 5) cos (x)

\frac{2 s + 9}{( s + 4 ) ( s + 6 )}

\int_{0}^{2} x^{3} e^{x^{2}} d x

\int_{- 1}^{3} ∣ x - 2 ∣ d x

What is the integral from 1 to 2 of 12x^22^{x^3} ?
The integral from 1 to 2 of 12x^22^{x^3} is (1016)/(ln(2))