What is the integral from 1 to sqrt(3 of)x2^{(x^2)} ?

The integral from 1 to sqrt(3 of)x2^{(x^2)} is 3/(ln(2))

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integral from 1 to sqrt(3 of)x2^{(x^2)}

\int_{1}^{3} x 2^{(x^{2})} d x

\frac{3}{ln ( 2 )}

Decimal Notation

4.32808 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{3} 2^{u - 1} d u

= \int_{1}^{3} 2^{u - 1} d u

Simplify

2^{u - 1} : 2^{u} \cdot 2^{- 1}

= \int_{1}^{3} 2^{u} \cdot 2^{- 1} d u

Take the constant out:

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

= 2^{- 1} \cdot \int_{1}^{3} 2^{u} d u

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

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

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

Compute the boundaries:

\frac{6}{ln ( 2 )}

= \frac{1}{2} \cdot \frac{6}{ln ( 2 )}

Simplify

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

\frac{d}{d x} (\frac{x ^{2} - 3 x}{x - 1})

h \to \infty lim (h^{2} - 3 h)

y^{^{''}} + 4 y^{^{'}} + 6 y = cos (x)

x \to \frac{1}{2} lim (2 x - 1)

\frac{d}{d x} (\frac{5 x + 2}{x + 1})

What is the integral from 1 to sqrt(3 of)x2^{(x^2)} ?
The integral from 1 to sqrt(3 of)x2^{(x^2)} is 3/(ln(2))